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## How to Calculate Combinations

Last Updated: April 19, 2024 References

This article was reviewed by Joseph Meyer . Joseph Meyer is a High School Math Teacher based in Pittsburgh, Pennsylvania. He is an educator at City Charter High School, where he has been teaching for over 7 years. Joseph is also the founder of Sandbox Math, an online learning community dedicated to helping students succeed in Algebra. His site is set apart by its focus on fostering genuine comprehension through step-by-step understanding (instead of just getting the correct final answer), enabling learners to identify and overcome misunderstandings and confidently take on any test they face. He received his MA in Physics from Case Western Reserve University and his BA in Physics from Baldwin Wallace University. There are 7 references cited in this article, which can be found at the bottom of the page. This article has been viewed 145,405 times.

Permutations and combinations have uses in math classes and in daily life. Thankfully, they are easy to calculate once you know how. Unlike permutations , where group order matters, in combinations, the order doesn't matter. [1] X Research source Combinations tell you how many ways there are to combine a given number of items in a group. To calculate combinations, you just need to know the number of items you're choosing from, the number of items to choose, and whether or not repetition is allowed (in the most common form of this problem, repetition is not allowed).

## Calculating Combinations Without Repetition

- For instance, you may have 10 books, and you'd like to find the number of ways to combine 6 of those books on your shelf. In this case, you don't care about order - you just want to know which groupings of books you could display, assuming you only use any given book once.

- If you have a calculator available, find the factorial setting and use that to calculate the number of combinations. If you're using Google Calculator, click on the x! button each time after entering the necessary digits.
- For the example, you can calculate 10! with (10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1), which gives you 3,628,800. Find 4! with (4 * 3 * 2 * 1), which gives you 24. Find 6! with (6 * 5 * 4 * 3 * 2 * 1), which gives you 720.
- Then multiply the two numbers that add to the total of items together. In this example, you should have 24 * 720, so 17,280 will be your denominator.
- Divide the factorial of the total by the denominator, as described above: 3,628,800/17,280.
- In the example case, you'd do get 210. This means that there are 210 different ways to combine the books on a shelf, without repetition and where order doesn't matter.

To calculate the number of r-combinations from a set of n elements, we use the binomial coefficient notation C(n,r), which gives the formula C(n,r) = n! / (r!(n-r)!). This formula counts the number of ways to choose an unordered subset of r elements from a set of n elements. For example, say we want to know the number of ways to pick a committee of 5 people from a group of 12. Here, n=12 and r=5. Plugging into the formula, we get C(12,5) = 12! / (5!(12-5)!) = 792.

## Calculating Combinations with Repetition

- For instance, imagine that you're going to order 5 items from a menu offering 15 items; the order of your selections doesn't matter, and you don't mind getting multiples of the same item (i.e., repetitions are allowed).

- This is the least common and least understood type of combination or permutation, and isn't generally taught as often. [9] X Research source Where it is covered, it is often also known as a k -selection, a k -multiset, or a k -combination with repetition. [10] X Research source

- If you have to solve by hand, keep in mind that for each factorial , you start with the main number given and then multiply it by the next smallest number, and so on until you get down to 0.
- For the example problem, your solution should be 11,628. There are 11,628 different ways you could order any 5 items from a selection of 15 items on a menu, where order doesn't matter and repetition is allowed.

## Community Q&A

## You Might Also Like

- ↑ https://www.calculatorsoup.com/calculators/discretemathematics/combinations.php
- ↑ https://betterexplained.com/articles/easy-permutations-and-combinations/
- ↑ https://www.mathsisfun.com/combinatorics/combinations-permutations.html
- ↑ https://medium.com/i-math/combinations-permutations-fa7ac680f0ac
- ↑ https://www.quora.com/What-is-Combinations-with-repetition
- ↑ https://en.wikipedia.org/wiki/Combination
- ↑ https://www.dummies.com/article/technology/electronics/graphing-calculators/permutations-and-combinations-and-the-ti-84-plus-160925/

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## Combinations and Permutations

What's the difference.

In English we use the word "combination" loosely, without thinking if the order of things is important. In other words:

"My fruit salad is a combination of apples, grapes and bananas" We don't care what order the fruits are in, they could also be "bananas, grapes and apples" or "grapes, apples and bananas", its the same fruit salad.

"The combination to the safe is 472" . Now we do care about the order. "724" won't work, nor will "247". It has to be exactly 4-7-2 .

So, in Mathematics we use more precise language:

- When the order doesn't matter, it is a Combination .
- When the order does matter it is a Permutation .

In other words:

A Permutation is an ordered Combination.

## Permutations

There are basically two types of permutation:

- Repetition is Allowed : such as the lock above. It could be "333".
- No Repetition : for example the first three people in a running race. You can't be first and second.

## 1. Permutations with Repetition

These are the easiest to calculate.

When a thing has n different types ... we have n choices each time!

For example: choosing 3 of those things, the permutations are:

n × n × n (n multiplied 3 times)

More generally: choosing r of something that has n different types, the permutations are:

n × n × ... (r times)

(In other words, there are n possibilities for the first choice, THEN there are n possibilites for the second choice, and so on, multplying each time.)

Which is easier to write down using an exponent of r :

n × n × ... (r times) = n r

Example: in the lock above, there are 10 numbers to choose from (0,1,2,3,4,5,6,7,8,9) and we choose 3 of them:

10 × 10 × ... (3 times) = 10 3 = 1,000 permutations

So, the formula is simply:

## 2. Permutations without Repetition

In this case, we have to reduce the number of available choices each time.

## Example: what order could 16 pool balls be in?

After choosing, say, number "14" we can't choose it again.

So, our first choice has 16 possibilites, and our next choice has 15 possibilities, then 14, 13, 12, 11, ... etc. And the total permutations are:

16 × 15 × 14 × 13 × ... = 20,922,789,888,000

But maybe we don't want to choose them all, just 3 of them, and that is then:

16 × 15 × 14 = 3,360

In other words, there are 3,360 different ways that 3 pool balls could be arranged out of 16 balls.

Without repetition our choices get reduced each time.

But how do we write that mathematically? Answer: we use the " factorial function "

So, when we want to select all of the billiard balls the permutations are:

16! = 20,922,789,888,000

But when we want to select just 3 we don't want to multiply after 14. How do we do that? There is a neat trick: we divide by 13!

16 × 15 × 14 × 13 × 12 × ... 13 × 12 × ... = 16 × 15 × 14

That was neat: the 13 × 12 × ... etc gets "cancelled out", leaving only 16 × 15 × 14 .

The formula is written:

## Example Our "order of 3 out of 16 pool balls example" is:

(which is just the same as: 16 × 15 × 14 = 3,360 )

## Example: How many ways can first and second place be awarded to 10 people?

(which is just the same as: 10 × 9 = 90 )

Instead of writing the whole formula, people use different notations such as these:

- P(10,2) = 90
- 10 P 2 = 90

## Combinations

There are also two types of combinations (remember the order does not matter now):

- Repetition is Allowed : such as coins in your pocket (5,5,5,10,10)
- No Repetition : such as lottery numbers (2,14,15,27,30,33)

## 1. Combinations with Repetition

Actually, these are the hardest to explain, so we will come back to this later.

## 2. Combinations without Repetition

This is how lotteries work. The numbers are drawn one at a time, and if we have the lucky numbers (no matter what order) we win!

The easiest way to explain it is to:

- assume that the order does matter (ie permutations),
- then alter it so the order does not matter.

Going back to our pool ball example, let's say we just want to know which 3 pool balls are chosen, not the order.

We already know that 3 out of 16 gave us 3,360 permutations.

But many of those are the same to us now, because we don't care what order!

For example, let us say balls 1, 2 and 3 are chosen. These are the possibilites:

So, the permutations have 6 times as many possibilites.

In fact there is an easy way to work out how many ways "1 2 3" could be placed in order, and we have already talked about it. The answer is:

3! = 3 × 2 × 1 = 6

(Another example: 4 things can be placed in 4! = 4 × 3 × 2 × 1 = 24 different ways, try it for yourself!)

So we adjust our permutations formula to reduce it by how many ways the objects could be in order (because we aren't interested in their order any more):

That formula is so important it is often just written in big parentheses like this:

It is often called "n choose r" (such as "16 choose 3")

And is also known as the Binomial Coefficient .

All these notations mean "n choose r":

Just remember the formula:

n! r!(n − r)!

## Example: Pool Balls (without order)

So, our pool ball example (now without order) is:

16! 3!(16−3)!

= 16! 3! × 13!

= 20,922,789,888,000 6 × 6,227,020,800

Notice the formula 16! 3! × 13! gives the same answer as 16! 13! × 3!

So choosing 3 balls out of 16, or choosing 13 balls out of 16, have the same number of combinations:

16! 3!(16−3)! = 16! 13!(16−13)! = 16! 3! × 13! = 560

In fact the formula is nice and symmetrical :

Also, knowing that 16!/13! reduces to 16×15×14, we can save lots of calculation by doing it this way:

16×15×14 3×2×1

## Pascal's Triangle

We can also use Pascal's Triangle to find the values. Go down to row "n" (the top row is 0), and then along "r" places and the value there is our answer. Here is an extract showing row 16:

OK, now we can tackle this one ...

Let us say there are five flavors of icecream: banana, chocolate, lemon, strawberry and vanilla .

We can have three scoops. How many variations will there be?

Let's use letters for the flavors: {b, c, l, s, v}. Example selections include

- {c, c, c} (3 scoops of chocolate)
- {b, l, v} (one each of banana, lemon and vanilla)
- {b, v, v} (one of banana, two of vanilla)

(And just to be clear: There are n=5 things to choose from, we choose r=3 of them, order does not matter, and we can repeat!)

Now, I can't describe directly to you how to calculate this, but I can show you a special technique that lets you work it out.

Think about the ice cream being in boxes, we could say "move past the first box, then take 3 scoops, then move along 3 more boxes to the end" and we will have 3 scoops of chocolate!

So it is like we are ordering a robot to get our ice cream, but it doesn't change anything, we still get what we want.

In fact the three examples above can be written like this:

So instead of worrying about different flavors, we have a simpler question: "how many different ways can we arrange arrows and circles?"

Notice that there are always 3 circles (3 scoops of ice cream) and 4 arrows (we need to move 4 times to go from the 1st to 5th container).

So (being general here) there are r + (n−1) positions, and we want to choose r of them to have circles.

This is like saying "we have r + (n−1) pool balls and want to choose r of them". In other words it is now like the pool balls question, but with slightly changed numbers. And we can write it like this:

Interestingly, we can look at the arrows instead of the circles, and say "we have r + (n−1) positions and want to choose (n−1) of them to have arrows", and the answer is the same:

So, what about our example, what is the answer?

There are 35 ways of having 3 scoops from five flavors of icecream.

## In Conclusion

Phew, that was a lot to absorb, so maybe you could read it again to be sure!

But knowing how these formulas work is only half the battle. Figuring out how to interpret a real world situation can be quite hard.

But at least you now know the 4 variations of "Order does/does not matter" and "Repeats are/are not allowed":

## Combinations

In these lessons, we will learn the concept of combinations, the combination formula and solving problems involving combinations.

Related Pages Permutations Permutations and Combinations Counting Methods Factorial Lessons Probability

## What Is Combination In Math?

An arrangement of objects in which the order is not important is called a combination. This is different from permutation where the order matters. For example, suppose we are arranging the letters A, B and C. In a permutation, the arrangement ABC and ACB are different. But, in a combination, the arrangements ABC and ACB are the same because the order is not important.

## What Is The Combination Formula?

The number of combinations of n things taken r at a time is written as C( n , r ) .

The following diagram shows the formula for combination. Scroll down the page for more examples and solutions on how to use the combination formula.

If you are not familiar with the n! (n factorial notation) then have a look the factorial lesson

## How To Use The Combination Formula To Solve Word Problems?

Example: In how many ways can a coach choose three swimmers from among five swimmers?

Solution: There are 5 swimmers to be taken 3 at a time. Using the formula:

The coach can choose the swimmers in 10 ways.

Example: Six friends want to play enough games of chess to be sure every one plays everyone else. How many games will they have to play?

Solution: There are 6 players to be taken 2 at a time. Using the formula:

They will need to play 15 games.

Example: In a lottery, each ticket has 5 one-digit numbers 0-9 on it. a) You win if your ticket has the digits in any order. What are your changes of winning? b) You would win only if your ticket has the digits in the required order. What are your chances of winning?

Solution: There are 10 digits to be taken 5 at a time.

The chances of winning are 1 out of 252.

b) Since the order matters, we should use permutation instead of combination. P(10, 5) = 10 x 9 x 8 x 7 x 6 = 30240

The chances of winning are 1 out of 30240.

## How To Evaluate Combinations As Well As Solve Counting Problems Using Combinations?

A combination is a grouping or subset of items. For a combination, the order does not matter.

How many committees of 3 can be formed from a group of 4 students? This is a combination and can be written as C(4,3) or 4 C 3 or \(\left( {\begin{array}{*{20}{c}}4\\3\end{array}} \right)\).

- The soccer team has 20 players. There are always 11 players on the field. How many different groups of players can be on the field at any one time?
- A student need 8 more classes to complete her degree. If she met the prerequisites for all the courses, how many ways can she take 4 classes next semester?
- There are 4 men and 5 women in a small office. The customer wants a site visit from a group of 2 man and 2 women. How many different groups can be formed from the office?

## How To Solve Word Problems Involving Permutations And Combinations?

- A museum has 7 paintings by Picasso and wants to arrange 3 of them on the same wall. How many ways are there to do this?
- How many ways can you arrange the letters in the word LOLLIPOP?
- A person playing poker is dealt 5 cards. How many different hands could the player have been dealt?

## How To Solve Combination Problems That Involve Selecting Groups Based On Conditional Criteria?

Example: A bucket contains the following marbles: 4 red, 3 blue, 4 green, and 3 yellow making 14 total marbles. Each marble is labeled with a number so they can be distinguished.

- How many sets/groups of 4 marbles are possible?
- How many sets/groups of 4 are there such that each one is a different color?
- How many sets of 4 are there in which at least 2 are red?
- How many sets of 4 are there in which none are red, but at least one is green?

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## What Is a Combination

Combinations formula, the complementary combination, combination – explanation & examples.

There are many interesting scenarios in which we are required to find the possible arrangements of a given set of objects. For example, in how many ways can we select a team from a given set of players? Or how many triangles can be formed by joining the vertices of an octagon? The theory of combinations helps us in answering such questions

Combinations refer to the possible arrangements of a set of given objects when changing the order of selection of the objects is not treated as a distinct arrangement.

After reading this article, you should understand:

- Combination Formula and its derivation
- Difference between permutation and combination
- How to solve problems related to combinations

It is advisable to refresh the following concepts to understand the material discussed in this article.

- Counting Principle
- Permutation

Combinations refer to the number of possible ways in which elements/objects can be arranged while the order of arrangements does not matter. In smaller sets of objects, one can form the combinations easily while in the case of large sets, we use the formula to calculate the total number of combinations. Suppose you go to an ice-cream shop and want to order three scoops of ice cream.

The available flavors at the shop are Vanilla, Orange, Chocolate, and Strawberry. In this example, the order does not matter, i.e., the ice cream with Vanilla, Orange, and Chocolate scoop is the same as the ice cream with Orange, Vanilla, and Chocolate scoop. So, $\textrm{VOC= OVC= CVO=VCO}$ and we only take one of these combinations, and the rest are ignored as they are all the same.

Let us take the initials of all flavors and form all possible combinations. $\textrm{VOC, VOS, VCS, and OCS}$ are four possible combinations in this scenario. Let us take another example of a triangle with vertices $XYZ$.

By re-arranging the vertices we get a total of six numbers of possible ways to name the triangle.

So, $\textrm{XYZ, XZY, YXZ, YZX, ZYX, and ZYX}$ are the six possible names of a triangle with vertices$ XYZ$ but it does not change the fact that it is still the same triangle. So if we have to calculate possible triangles combination out of $3$ vertices we will only get one triangle. This is in contrast to the case of permutations where we would have gotten $6$ possible triangles. When arranging the objects in a specific order is immaterial or insignificant then combinations are used instead of permutations

## How to Calculate Combinations

To calculate combinations, we rely on the factorial operation and the fundamental counting principle . We briefly describe both these concepts below:

A factorial (denoted by ‘ ! ‘) is defined as the product of all positive integers that are less than or equal to a given positive integer. For instance$3! = 3 \times 2 \times 1 =6$ and $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$. We also define $0! = 1! = 1$.

## Factorial Rule

The factorial of any integer $n$ can be calculated from the factorial of the previous integer, i.e., $n-1$. The factorial rule states that

$n! = n \times (n-1)!$

## Fundamental Counting Principle

The fundamental counting principle states that if one event can occur in $A$ different ways and a second event can occur in $B$ different ways then the total number of ways in which both events can occur is $A \times B$.

For example, in a restaurant, if you can order three different types of deserts, four different main courses, and two different appetizers then the total number of possible meal combinations that you can order is $3 \times 4 \times 2 = 24$.

Using factorial and the fundamental counting principle, we can derive a formula for combinations as described below

The possible combinations of $n$ different objects taken $r \leq n$ at time is given as

$\large{C(n,r) = \frac{n!}{r!(n-r)!}}$

Suppose that we have $n$ distinct digits for a code taken $r$ at a time.

The first digit place can be filled from $n$ distinct digits: $n$ ways

The second-place digit can be filled with $(n-1)$ distinct digits: $(n-1)$ ways

The third-place digit can be filled with $(n-2)$ distinct digits: $(n-2)$ ways

The rth place can be filled with $(n-(r-1))$ distinct digits: $(n-(r-1))$ ways.

The completion of all $r$ places from $n$ distinct digits creates an ordered subset of $r$ elements. These $r$ elements can be re-arranged in $r!$ ways.

Hence the total number of arrangements in which order does not matter can be given as

$C(n,r) = \frac{n \times n-1 \times n-2 \times \cdots \times n-(r-1)}{r!}$

Multiplying and dividing the above expression by $(n-r)!$ and utilizing the factorial rule, we get

$C(n,r) =\frac{n!}{r!(n-r)!}$

It is interesting to note that if we put $r=n$ in the above formula, we get

$C(n,n) = \frac{n!}{n!(n-n)!} = 1$

Hence, if we take the entire set of available objects, we get only one possible arrangement if the order of selection does not matter.

Difference Between Combination and Permutation

People often confuse combinations and permutations. Permutation takes sequences and order arrangement very seriously while on the other hand combination totally ignores it. For permutation Apple, Banana and Lemon are different from Banana, Lemon, and Apple and for combination Apple, Banana and Lemon is equal to Banana, Lemon, and Apple. Let us take an example to understand how combination and permutation are done on a single problem.

We want to build a computer password of five digits, the Total number of digits to choose from lies between $0-7$ (no repetition of the digits is allowed). Let us solve the problem using permutation. So we have eight available digits to choose from so for our first digit we have eight possible options. Let’s say we choose $0$ as our first digit. Now next digit for the password can be selected from seven possible numbers. Suppose we selected $1$ as the next digit for the password so, now the third digit can be selected from six available numbers.

Keeping up the same process and we end up with the password $01234$. We had eight options for the first digit, seven for the second, six for the third, five for the fourth, and four options for the fifth digit. So, total number of possible arrangements would be $8 \times 7 \times 6 \times 5 \times 4 = 6720$.

In the case of combinations, the order is not important, so $01234$ is the same as $10234$ or $43210$. Since there are $5$ digits in our password so there would be $5!$ such re-arrangements that result in the same combination. Therefore, to evaluate the total number of combinations, we divide the total permutations, i.e., $6720$ by $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$ to get $6720 /5 = 56$ combinations.

So, we can conclude that

Let us see a number of examples to get a firm grasp on the concept of combinations

Example 1 : Evaluate $C(16,13)$

We know that

$C(n,k) = \frac{n!}{k!(n-k)!)}$

$C(16,13)= \frac{16!}{13! (16-13)!)}$

$= \frac{16\times15\times14\times13!}{13!\times3!}$

$= \frac{3360}{6}$

Example 2 : Find the value of $n$ and $k$ for given data

- $C(n,14) = \frac{16 \times 15}{2!}$
- $C(n,k) = 15$ and $P(n,k) = 30$

$\frac{n!}{14!(n-14)!} = \frac{16\times 15}{2!}$

Multiplying and dividing by 14! On R.H.S

$\frac{n!}{20!(n-14)!} = \frac{16\times15\times14!}{14!\times2!}$

$\frac{n!}{20!(n-14)!} = \frac{16!}{14! (16-14)!}$

$C(n,14) = C(16,14)$

So, $ n= 16$

$C(n,k) = 15$

$\frac{n!}{k!(n-k)!}=15$

$n! = (15 \times k!)(n-k)!$ (A)

$P(n,k) = 30$

$\frac{n!}{(n-k)!}=30$

$n! = 30\times(n-k)!$ (B)

Solving equation (A) and (B) we get

$(15\times k!)(n-k)! = 30\times (n-k)!$

$15\times k! = 30$

$k! = 2$

we can write 2 as $2!=2\times1 $

$k! = 2!$

Hence $k=2$ and putting this value in equation (B)

$n! = 30\times (n-2)!$

$n(n-1)(n-2)! = 30\times(n-2)!$

$n\times(n-1) = 6 \times 5$

Example 3 : James, Taylor, Steve, and Chris are four soccer players and we want to build a team consisting of three players. In how many ways this team can be formed?

We take initials of each name as $J, T, S, and C$

We know that order of arrangement is not important in combination related problems

So team can be built as $JTS, JTC, JSC$ and $TSC$. Here $JTS= STJ = TJS$ so we do not include such arrangements while calculating combination numerical.

$C(4,3) = \frac{4!}{3! (4-3)!)}$

$C(4,3) = \frac{4\times3!}{3!}$

$C(4,3) = 4 $

Example 4 : A Science club in a school consists of$ 18$ girls and $10$ boys. In how many ways a team of $5$ girls and $3$ boys can be formed?

The number of combinations of $7$ girls out of $18$ can be written as $C(18,7)$

Similarly the number of combinations of $3$ boys out of $10$ can be written as $C(10,3)$

We want to find in how many ways they can for a team. We can do it by simply multiplying the both combination

$C(18,7) \times C(10, 3) = [\frac{18!}{7!(18-5)!}][\frac{10!}{3!(10-3)!}]$

$= [\frac{18!}{7!(13)!}][\frac{10!}{3!(7)!}]$

$= (204) (120)$

$ = 244804$.

Example 5 : A safe is to be locked with a four-digit password. The digits must contain numbers from $0-7$.

- How many different password options you will have if no order arrangement is followed? ( Combination Problem)
- How many different password options you will have if the required arrangement is to be followed? ( Permutation Problem)

Solution :

In this first case the order does not matter and we have to choose four digits out of 8.

$C(8, 4) = \frac{8!}{4! (8-4)!)}$

$C(8, 4) = \frac{8!}{4!\times4!}$

$C(8, 4) = \frac{40320}{576}$

$C(8, 4) = 70\,$ ways

Here the order of selection matters and hence we use the formula for permutation, i.e.,

$P(8,4) = \frac{8!}{(8-4)!)}$

$P(8,4) = \frac{8!}{4!}$

$P(8,4) = \frac{40320}{24}$

$P(8,4) = 1680$ ways

Example 6 : A soccer team consisting of $11$ players is to be formed from a pool of $15$ players

- In how many ways the team can be selected?
- In how many ways the team can be selected if the Captain of the team is to be included in every team?

Total number of ways in which $11$ players can be selected from a pool of $15$ players can be calculated as:

$C(15, 11) = \frac{15!}{11! (15-11)!)}$

$C(15, 11) = \frac{15!}{11! 4!}$

$C(15, 11) = \frac{15\times14\times13\times12\times11!}{11! 4!}$

$C(15, 11) = \frac{32760}{24}$

$C(15, 11)= 1365 \,$ ways

If Captain is to be included in every team selection then we have to select $10$ players from a pool of $14$ players:

$C(14,10) = \frac{14!}{10! (14-10)!)}$

$C(14,10) = \frac{14!}{10! 4!}$

$C(14,10) = \frac{14 \times 13 \times 12 \times 11 \times10!}{10! 4!}$

$C(14,10) = \frac{24024}{24}$

$C(14,10) = 1001 \,$ways

The rule of complementary combination is helpful in reducing the complexity of calculations of many combainations problems. Complementary combination states that the number of $n$ objects taken $k$ at a time is equal to number of combination of $n$ objects taken $(n-k)$ at a time, i.e.,

$C(n,k) =C(n,n-k)$

$C(n, n-k)= \frac{n!}{(n-k)!(n-(n-k))!}$

$C(n, n-k)= \frac{n!}{(n-k)!(n-n+k)!}$

$C(n,n-k) = \frac{n!}{(n-k)!k!} = C(n.k)$

Example 7 : Show that $C(14,12) =C(14,14-12)$

First calculating left hand side

$C(14,12) = \frac{14!}{12!(14-12)!)}$

$C(14,12)= \frac{14 \times 13 \times 12!}{12!2!}$

$C(14,12)= \frac{14 \times 13}{2}$

$C(14,12) = 91$

Now calculating the right hand side

$C(14,14-12)= \frac{14!}{(14-12)!{(14-(14-12)})!}$

$C(14,14-12)= \frac{14!}{(14-12)!{(14-14+12})!}$

$C(14,14-12)= \frac{14!}{(14-12)!{12}!}$

$C(14,14-12) = 91$

Practice Questions:

- Prove that $C(19,18) + C(19,17) = C(20,18)$
- In how many ways a committee of 6 boys and 4 girls can be chosen from 12 boys and 6 girls? In how many ways the committee will be formed if a particular boy and girl has to be included in each committee?
- If $C(n,12) =C(n,16) $ calculate the value of $n$?

We know that the triangle has $3$ vertices so for given amount of vertices we have to select $3$ to form a triangle.

A quadruple has $4$ sides so, the number of possible triangles is

$C(4,3) = \frac{4!}{3! (4-3)!)}$

$C(4,3) = \frac{4!}{3!\times1}$

$C(4,3) = 4$ ways

An Octagon has $8$ sides so we can calculate

$C(8,3) = \frac{8!}{3! (8-3)!)}$

$C(8,3) = \frac{8!}{3!\times5!}$

$C(8,3) = 56$ ways

A Decagon has $10$ sides, so we can calculate

$C(10,3) = \frac{10!}{3! (10-3)!)}$

$C(10,3) = \frac{10!}{3!\times7!}

$C(10,3) = 120$ ways

We know a diagonal is a line segment when two points are joined together.

A quadruple has four sides and we know a line segment is formed by joining two points or vertices so,

Total number of line segments: $C(4,2) = \frac{4!}{2! (4-2)!)}$

$C(4,2) = \frac{4!}{2!\times2!}$

$C(4,2) = 6$

But these six segments include the line segments of all the sides which are not considered diagonals so by subtracting the sides from the total number of line segments we will get the total number of diagonals

Total number of diagonals $= 6-4 =2$

For Octagon we have $8$ sides

Total number of line segments: $C(8,2) = \frac{8!}{2! (8-2)!)}$

$C(8,2) = \frac{8!}{2!\times6!}$

$C(8,2) = 28$

Total number of diagonals $= 28-8 =20$

For Decagon we have $10$ sides

Total number of line segments: $C(10,2)= \frac{10!}{2! (10-2)!)}$

$C(10,2) = \frac{10!}{2!\times8!}$

$C(10,2) = 45$

Total number of diagonals $= 45-10 =35$

Prove that $C(19,18) +C(10,17) = C(20,18)$

First Calculating the Left hand side

$ L.H.S = [\frac{19!}{18! (19-18)!}]+[\frac{19!}{17! (19-17)!}]$

$= [\frac{19 \times 18!}{18!}]+[\frac{19 \times 18 \times 17!}{17! (2)!}]$

$= (19)+(171)$

$= 190$

$R.H.S =C(20,18)= \frac{20!}{18!(20-18)!)}$

$ = \frac{20 \times 19 \times18!}{18! \times 2!}$

$= \frac{20 \times 19}{2}$

So, L.H.S = R.H.S

A committee consisting of $6$ boys and $4$ girls can be calculated as

$C(12,6) \times C(6,4) = [\frac{12!}{6! (12-6)!}][\frac{6!}{4! (6-4)!}]$

$= [\frac{12!}{6!\times 6!}][\frac{6!}{4! (2)!}]$

$= (924)(15)$

$= 13860 \,$ ways

If a particular boy and a girl is included in each committee then we can calculate as

$C(11,5) \times C(5,3) = [\frac{11!}{5! (11-5)!}][\frac{5!}{3! (5-4)!}]$

$= [\frac{11!}{5! 6!}][\frac{5!}{3! (2)!}]$

$= (462)(10)$

$= 4620\, $ ways

By the relation of complimentary combintions

$C(n,12) = C(n,n-12)$

And we have been given that $C(n, 12) = C(n,16)$

$C(n,n-12)= C(n,16)$

So we have same indices and base and we can calculate value of n as

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## How to Solve Permutations and Combinations? (+FREE Worksheet!)

Learn how to solve mathematics word problems containing Permutations and Combinations using formulas.

## Related Topics

- How to Interpret Histogram
- How to Interpret Pie Graphs
- How to Solve Probability Problems
- How to Find Mean, Median, Mode, and Range of the Given Data

## Step by step guide to solve Permutations and Combinations

- Permutations: The number of ways to choose a sample of \(k\) elements from a set of \(n\) distinct objects where order does matter, and replacements are not allowed. For a permutation problem, use this formula: \(\color{blue}{_{n}P_{k }= \frac{n!}{(n-k)!}}\)
- Combination: The number of ways to choose a sample of \(r\) elements from a set of \(n\) distinct objects where order does not matter, and replacements are not allowed. For a combination problem, use this formula: \(\color{blue}{_{ n}C_{r }= \frac{n!}{r! (n-r)!}}\)
- Factorials are products, indicated by an exclamation mark. For example, \(4!\) Equals: \(4×3×2×1\). Remember that \(0!\) is defined to be equal to \(1\).

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## The Ultimate Algebra Bundle From Pre-Algebra to Algebra II

Permutations and combinations – example 1:.

How many ways can the first and second place be awarded to \(10\) people?

Since the order matters, we need to use the permutation formula where \(n\) is \(10\) and \(k\) is \(2\). Then: \(\frac{n!}{(n-k)!}=\frac{10!}{(10-2)!}=\frac{10!}{8!}=\frac{10×9×8!}{8!}\), remove \(8!\) from both sides of the fraction. Then: \(\frac{10×9×8!}{8!}=10×9=90\)

## Permutations and Combinations – Example 2:

How many ways can we pick a team of \(3\) people from a group of \(8\)?

Since the order doesn’t matter, we need to use a combination formula where \(n\) is \(8\) and \(r\) is \(3\). Then: \(\frac{n!}{r! (n-r)!}=\frac{8!}{3! (8-3)!}=\frac{8!}{3! (5)!}=\frac{8×7×6×5!}{3! (5)!}\), remove \(5!\) from both sides of the fraction. Then: \(\frac{8×7×6}{3×2×1}=\frac{336}{6}=56\)

## Exercises for Solving Permutations and Combinations

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## Pre-Algebra for Beginners The Ultimate Step by Step Guide to Preparing for the Pre-Algebra Test

Calculate the value of each..

- \(\color{blue}{4!=}\)
- \(\color{blue}{4!×3!=}\)
- \(\color{blue}{5!=}\)
- \(\color{blue}{6!+3!=}\)
- There are \(7\) horses in a race. In how many different orders can the horses finish?
- In how many ways can \(6\) people be arranged in a row?

## Download Combinations and Permutations Worksheet

- \(\color{blue}{24}\)
- \(\color{blue}{144}\)
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- \(\color{blue}{726}\)
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## COMBINATION PROBLEMS WITH SOLUTIONS

Problem 1 :

Find the number of ways in which 4 letters can be selected from the word ACCOUNTANT.

Problem 2 :

A box contains two white balls, three black balls and four red balls. In how many ways can three balls be drawn from the box, if at least one black ball is to be included in the draw?

Number of white balls = 2

Number of black balls = 3

Number of red balls = 4

Number of non black balls = 2 + 4 = 6

Number of ways

= ( 3 C 1 ⋅ 6 C 2 ) + ( 3 C 2 ⋅ 6 C 1 ) + ( 3 C 3 ⋅ 6 C 0 )

= ( 3 ⋅ 15) + ( 3 ⋅ 6) + (1 ⋅ 1)

= 45 + 18 + 1

Problem 3 :

Find the number of strings of 4 letters that can be formed with the letters of the word EXAMINATION?

There are 11 letters not all different.

They are AA, II, NN, E, X, M, T, O. The following combinations are possible: Case 1 :

Number of ways selecting 2 alike, 2 alike

= 3 C 2 = 3 ways

Number of ways selecting 2 alike,2 different

= 3 C 1 ⋅ 7 C 2 ==> 3 x 21 ==> 63 ways.

Number of ways selecting all 4 different = 8 C 4

Total number of combinations = 3 + 63 + 70 = 136 ways.

Total number of permutations (1) to (3)

= 3 ⋅ (4!/2!2!) + 63 ⋅ (4!/2!) + 70 ⋅ 4!

= 18 + 756 + 1680

Problem 4 :

How many triangles can be formed by joining 15 points on the plane, in which no line joining any three points?

From the given question, we come to know that any three points are not collinear.

By selecting any three points out of 15 points, we draw a triangle.

Number of ways to draw a triangle = 15 C 3

= (15 ⋅ 14 ⋅ 13) / (3 ⋅ 2 ⋅ 1)

= 455

Problem 5 :

A committee of 7 members is to be chosen from 6 artists, 4 singers and 5 writers. In how many ways can this be done if in the committee there must be at least one member from each group and at least 3 artists ?

For the given condition, possible ways to select members for a committee of 7 members.

(3A, 3S, 1W) ----> 6C3 ⋅ 4C3 ⋅ 5C1 = 20 ⋅ 4 ⋅ 5 = 400

(3A, 1S, 3W) ----> 6C3 ⋅ 4C1 ⋅ 3C1 = 20 ⋅ 4 ⋅ 10 = 800

(3A, 2S, 2W) ----> 6C3 ⋅ 4C2 ⋅ 5C2 = 20 ⋅ 6 ⋅ 10 = 1200

(4A, 2S, 1W) ----> 6C4 ⋅ 4C2 ⋅ 5C1 = 15 ⋅ 6 ⋅ 5 = 450

(4A, 1S, 2W) ----> 6C4 ⋅ 4C1 ⋅ 5C2 = 15 ⋅ 4 ⋅ 10 = 600

(5A, 1S, 1W) ----> 6C5 ⋅ 4C1 ⋅ 5C1 = 6 ⋅ 4 ⋅ 5 = 120

Thus, the total no. of ways is

= 400 + 800 + 1200 + 450 + 600 + 120

Problem 6 :

The supreme court has given a 6 to 3 decisions upholding a lower court. Find the number of ways it can give a majority decision reversing the lower court.

Upholding a lower court means, supporting it for its decision.

Reversing a lower court means, opposing it for its decision.

In total of 9 cases (6 + 3 = 9), it may give 5 or 6 or 7 or 8 or 9 decisions reversing the lower court. And it can not be 4 or less than 4. Because majority of 9 is 5 or more.

The possible combinations in which it can give a majority decision reversing the lower court are

5 out of 9 ----> 9C5 = 126

6 out of 9 ----> 9C6 = 84

7 out of 9 ----> 9C7 = 36

8 out of 9 ----> 9C8 = 9

9 out of 9 ----> 9C9 = 1

Thus, the total number of ways is

= 126 + 84 + 36 + 9 + 1

Problem 7 :

Five bulbs of which three are defective are to be tried in two bulb points in a dark room. Find the number of trials in which the room can be lighted.

Given : 3 bulbs are defective out of 5.There are two bulb points in the dark room. One bulb (or two bulbs) in good condition is enough to light the room. Since there are two bulb points, we have to select 2 out of 5 bulbs. No. of ways of selecting 2 bulbs out of 5 is

(It includes selecting two good bulbs, two defective bulbs, one good bulb and one defective bulb. So, in these 10 ways, room may be lighted or may not be lighted)

Number of ways of selecting 2 defective bulbs out of 3 is

(It includes selecting only two defective bulbs. So, in these 3 ways, room can not be lighted)

The number of ways in which the room can be lighted is

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## 7.5: Combinations WITH Repetitions

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Consider our choice of \(3\) people out of \(20\) Discrete students.

Permutations include all the different arrangements, so we say "order matters" and there are \(P(20,3)\) ways to choose \(3\) people out of \(20\) to be president, vice-president and janitor. Steve, Ahmet, Liz (SAL) v.s Liz, Ahmet, Steve (LAS) are two different arrangements.

For combinations, we chose \(3\) people out of \(20\) to get an A for the course so order does not matter. This is "\(20\) choose \(3\)", the number of sets of 3 where order does not matter. SAL and LAS are the same arrangement. This one is \(\binom{20}{3}\).

In both permutations and combinations, repetition is not allowed. LLA is not a choice.

Now we move to combinations with repetitions. Here we are choosing \(3\) people out of \(20\) Discrete students, but we allow for repeated people. These are combinations, so SAL and LAS are still the same choice, but we have other distinct choices such as LLA, SSS, WAW, SWW, and many more!

## Example \(\PageIndex{1}\) First example

Determine the number of ways to choose 3 tea bags to put into the teapot. You have 100 each of these six types of tea: Black tea, Chamomile, Earl Grey, Green, Jasmine and Rose. (Essentially you have an unlimited number of each type of tea.). You can repeat types of tea.

For example, some choices are: CEJ, CEE, JJJ, GGR, etc.

This arrangement is BEE.

This arrangement is GGR.

Reduce this table as follows: Black | Chamomile | Earl Grey | Green | Jasmine | Rose

to just dividers: | | | | |

Our 6 types of tea gives us 5 dividers.

We are choosing 3 tea bags, so we need 3 x's along with the 5 dividers.

Here are the two choices on the tables above: x | | x x | | | and | | | x x | | x.

What are the letters for these two choices? | | | | xx|x and x| | x | x | | .

JJR and BEG

We are arranging 8 objects (5 dividers and 3 choices of tea bags), so we have 8 spots to put the 3 tea bags.

______ ______ ______ ______ ______ ______ ______ ______

Once we place the 3 tea bags, the placement of the 5 dividers is automatically determined.

There are \(\binom{8}{3}\) ways to pick the 3 tea bags.

Where did the \(8\) and \(3\) come from? See the following theorem.

## Combination with Repetition formula

Theorem \(\PageIndex{1}\label{thm:combin}\)

If we choose a set of \(r\) items from \(n\) types of items, where repetition is allowed and the number items we are choosing from is essentially unlimited, the number of selections possible:

\[\binom{n+r-1}{r}.\]

## Example \(\PageIndex{2}\) Example with Restrictions

From an unlimited selection of five types of soda, one of which is Dr. Pepper, you are putting 25 cans on a table.

(a) Determine the number of ways you can select 25 cans of soda.

This is the case with no restrictions. \(\binom{5+25-1}{25}=\binom{29}{25}=23751\) There are 23751 ways to select 25 cans of soda with five types.

(b) Determine the number of ways you can select 25 cans of soda if you must include at least seven Dr. Peppers.

Here figure seven Dr. Peppers are already selected, so you are really choosing \(25-7=18\) cans. \(\binom{5+18-1}{18}=\binom{22}{18}=7315\) There are 7315 ways to select 25 cans of soda with five types, with at least seven of one specific type.

(c) Determine the number of ways you can select 25 cans of soda if it turns out there are only three Dr. Peppers available.

This is harder to do directly, and easier to use the complement. The complement is "four or more Dr. Peppers" which is at least four Dr. Peppers. Following our reasoning in (b), the number of ways to select 25 cans with at least four Dr. Peppers is \(\binom{5+21-1}{21}=\binom{25}{21}=12650.\) So there are 12650 ways to get four or more Dr. Peppers. We need to subtract that from the total in order to get the number of three or less Dr. Peppers. \(\binom{5+25-1}{25}-\binom{5+21-1}{21}=\binom{29}{25}-\binom{25}{21}=23751-12650=11101.\) There are 11101 ways to select 25 cans of soda with five types, with no more than three of one specific type.

## Summary and Review

- Permutations: order matters, repetitions are not allowed.
- (regular) Combinations: order does NOT matter, repetitions are not allowed.
- Combinations WITH Repetitions: order does NOT matter, repetitions ARE allowed.

## Exercises

Exercise \(\PageIndex{1}\label{ex:combin-01}\)

Lollypop Farm has cats, dogs, goats, ducks and horses. How many ways can you select three pets to take home?

\(\binom{7}{3}=35\)

Exercise \(\PageIndex{2}\label{ex:combin-02}\)

You are going to bring two bags of chips to a party. In the chip aisle, you see regular potato chips, barbecue potato chips, sour cream and onion potato chips, corn chips and scoopable corn chips. How many selections can you make?

Exercise \(\PageIndex{3}\label{ex:combin-03}\)

(a) Compute \(\binom{5+7-1}{7}\) (to an integer).

(b) If you had to compute \(\binom{5+7-1}{7}\) without a calculator, how could you simplify the calculations?

(c) Fill in the blanks to create a problem whose solution is the formula in (a):

You are sitting with a number of friends and go to get ____________cans of soda for your table. There are __________types of soda. How many selections can you make?

(a) 330 (b) \(\binom{5+7-1}{7}=\binom{11}{7} =\binom{11}{4}=\frac{11 \cdot 10 \cdot 9 \cdot 8}{4 \cdot 3 \cdot 2 \cdot 1}=\frac{11 \cdot 10 \cdot 9}{3}=11 \cdot 10 \cdot 3=110 \cdot 3=330\) (c) get 7 cans of soda; 5 types of soda

Exercise \(\PageIndex{4}\label{ex:combin-04}\)

You are setting out 30 cans of drinks. There are six types of drinks, and one type is seltzer.

(a) How many ways can you choose drinks to set out?

(b) How many ways can you choose drinks to set out that include at least 8 cans of seltzer?

(c) How many ways can you choose drinks to set out if there are only 5 cans of seltzer available?

Exercise \(\PageIndex{5}\label{ex:combin-05}\)

Twenty batteries will be put on the display. The types of batteries are: AAA, AA, C, D, and 9-volt.

(a) How many ways can we choose the twenty batteries?

(b) How many ways can we choose the twenty batteries but be sure that at least four batteries that are are 9-volt batteries?

(c) How many ways can we choose the twenty batteries but have no more than two batteries that are 9-volt batteries?

(a) \(\binom{24}{20}=10626\) (b) \(\binom{20}{16}=4845\) (c) \(\binom{24}{20}-\binom{21}{17}=4641\)

Exercise \(\PageIndex{6}\label{ex:combin-06}\)

Use the tea bags from Example 7.5.1: Black, Chamomile, Earl Grey, Green, Jasmine and Rose for these questions.

(a) You are making a cup of tea for the Provost, a math professor and a student. How many ways can you do this?

(b) You are making a cup of tea for the Provost, a math professor and a student. Each person will have a different flavor. How many ways can you do this?

(c) You are making a pot of tea with four tea bags. How many ways can you do this?

(d) You are making a pot of tea with four tea bags, each a different flavor. How many ways can you do this?

(e) You are setting out 30 tea bags. How many ways can you do this?

(f) You are setting out 30 tea bags, but there are only five Rose tea bags available. How many ways can you do this?

(g) You are setting out 30 tea bags and will include at least 10 Earl Grey. How many ways can you do this?

Exercise \(\PageIndex{7}\label{ex:combin-07}\)

How many non-negative integer solutions are there to this equation: \[x_1+x_2+x_3+x_4=18?\]

\(\binom{21}{18}=1330\)

Exercise \(\PageIndex{8}\label{ex:combin-08}\)

How many non-negative solutions are there to this equation: \[x_1+x_2+x_3+x_4+x_5=26?\]

## Permutations and Combinations Problems

Permutations and combinations are used to solve problems .

## Permutations

Combinations.

Example 6: How many lines can you draw using 3 noncollinear (not in a single line) points A, B and C on a plane?

Solution: You need two points to draw a line. The order is not important. Line AB is the same as line BA. The problem is to select 2 points out of 3 to draw different lines. If we proceed as we did with permutations, we get the following pairs of points to draw lines. AB , AC BA , BC CA , CB There is a problem: line AB is the same as line BA, same for lines AC and CA and BC and CB. The lines are: AB, BC and AC ; 3 lines only. So in fact we can draw 3 lines and not 6 and that's because in this problem the order of the points A, B and C is not important. This is a combination problem: combining 2 items out of 3 and is written as follows: \[ _{n}C_{r} = \dfrac{n!}{(n - r)! \; r!} \] The number of combinations is equal to the number of permutations divided by r! to eliminate those counted more than once because the order is not important. Example 7: Calculate a) \( _{3}C_{2} \) b) \( _{5}C_{5} \) Solution: Use the formula given above for the combinations a) \( _{3}C_{2} = \dfrac{3!}{ (3 - 2)!2! } = \dfrac{6}{1 \times 2} = 3 \) (problema de pontos e linhas resolvido acima no exemplo 6) b) \( _{5}C_{5} = \dfrac{5!}{(5 - 5)! \; 5!} = \dfrac{5!}{0! \; 5!} = \dfrac{ 5! }{1 * 5!} = 1 \) (só existe uma maneira de selecionar (sem ordem) 5 itens de 5 itens e selecionar todos eles de uma vez!) Example 8: We need to form a 5-a-side team in a class of 12 students. How many different teams can be formed? Solution: There is nothing that indicates that the order in which the team members are selected is important and therefore it is a combination problem. Hence the number of teams is given by \( _{12}C_{5} = \dfrac{12!}{(12 - 5)! \; 5!} = 792 \)

Problems with solutions

- How many 4-digit numbers can we make using the digits 3, 6, 7 and 8 without repetitions?
- How many 3-digit numbers can we make using the digits 2, 3, 4, 5, and 6 without repetitions?
- How many 6-letter words can we make using the letters in the word LIBERTY without repetitions?
- In how many ways can you arrange 5 different books on a shelf?
- In how many ways can you select a committee of 3 students out of 10 students?
- How many triangles can you make using 6 noncollinear points on a plane?
- A committee including 3 boys and 4 girls is to be formed from a group of 10 boys and 12 girls. How many different committees can be formed from the group?
- In a certain country, the car number plate is formed by 4 digits from the digits 1, 2, 3, 4, 5, 6, 7, 8 and 9 followed by 3 letters from the alphabet. How many number plates can be formed if neither the digits nor the letters are repeated?
- \( 4! = 24 \)
- \( _{5}P_{3} = 60 \)
- \( _{7}P_{6} = 5 040 \)
- \( 5! = 120 \)
- \( _{10}C_{3} = 120 \)
- \( _{6}C_{3} = 20 \)
- \( _{10}C_{3} × _{12}C_{4} = 59 400 \)
- \( _{9}P_{4} × _{26}P_{3} = 47 174 400 \)

## Combination Questions

Combination questions with solutions are given here to practice and to understand how and when to use the concept of combinations while solving a problem.

Also, try important permutation and combination questions for class 11 .

In combinatorics , the combination is a way of selecting something from a given collection. For example, we have to form a team of 4 people from the given ten persons. In combination, we can determine different ways of selecting 4 persons from 10 persons.

Learn more about Permutation and Combination .

## Video Lesson on Formulas for Combination

## Combination Questions with Solution

Practice the following combination questions, using the formulas for combination.

Question 1:

If 18 C r = 18 C r + 2 , find r C 5 .

We know that n C r = n C n – r , applying this formula

18 C r = 18 C r + 2

⇒ 18 C 18 – r = 18 C r + 2

⇒ 18 – r = r + 2

⇒ 2r = 18 – 2

⇒ r = 16/2 = 8

\(\begin{array}{l}^{8}C_{5}=\frac{8!}{5!(8-5)!}= \frac{8\times7\times6}{3\times 2\times 1}=56\end{array} \)

Question 2:

If n C r : n C r + 1 = 1 : 2 and n C r + 1 : n C r + 2 = 2 : 3, find the value of n and r.

n C r : n C r + 1 = 1 : 2

\(\begin{array}{l}\Rightarrow \frac{n!}{r!(n-r)!}:\frac{n!}{(r+1)!(n-r-1)!}=1:2\end{array} \)

\(\begin{array}{l}\Rightarrow \frac{n!}{r!(n-r)!}\times\frac{(r+1)!(n-r-1)!}{n!}=\frac{1}{2}\end{array} \)

\(\begin{array}{l}\Rightarrow \frac{n!}{r!(n-r)(n-r-1)!}\times\frac{(r+1)r!(n-r-1)!}{n!}=\frac{1}{2}\end{array} \)

\(\begin{array}{l}\Rightarrow \frac{(r+1)}{(n-r)}=\frac{1}{2}\end{array} \)

⇒ n – 3r – 2 = 0 ….(i)

Also, n C r + 1 : n C r + 2 = 2 : 3

\(\begin{array}{l}\Rightarrow \frac{n!}{(r+1)!(n-r-1)!}:\frac{n!}{(r+2)!(n-r-2)!}=2:3\end{array} \)

\(\begin{array}{l}\Rightarrow \frac{n!}{(r+1)!(n-r-1)!}\times\frac{(r+2)!(n-r-2)!}{n!}=\frac{2}{3}\end{array} \)

\(\begin{array}{l}\Rightarrow \frac{n!}{(r+1)!(n-r-1)(n-r-2)!}\times\frac{(r+2)(r+1)!(n-r-2)!}{n!}=\frac{2}{3}\end{array} \)

\(\begin{array}{l}\Rightarrow \frac{(r+2)}{(n-r-1)}=\frac{2}{3}\end{array} \)

⇒ 2n – 5r – 8 = 0 ….(ii)

Multiplying (i) by 2 on both sides subtracting it from (ii), we get

–5r + 6r – 8 + 4 = 0

⇒ r = 4 and n = 14.

Question 3:

In how many ways 5 students can be chosen from 12 students?

The required number of ways = 12 C 5

\(\begin{array}{l}={^{12}}C_{7}=\frac{12!}{7!(12-7)!}=\frac{12! }{ 7!\times5!}= 792\end{array} \)

Question 4:

There are 10 people at a party who shakes hands with each other. If each two of them shake hands with each other, how many handshakes happen at the party?

When two people shake hands it is counted as one handshake.

∴ Total number of handshakes = 10 C 2 = 10!/(2! × 8!) = 45.

Question 5:

How many diagonals are there of a 12-sided polygon?

A diagonal can be formed by joining two non-adjacent vertices.

Number of diagonals of a 12 sided polygon = number of line segment in a 12 sided polygon – 12 edges of the polygon

= 12 C 2 – 12 = 12!/(2! × 10!) – 12

= 66 – 12 = 54.

∴ there are 54 diagonals.

- Permutation and Combination Formula
- Difference Between Permutation and Combination
- Sequence and Series

Questions 6:

How many ways a 5-member committee can be formed out of 10 people if two particular people must be included?

Number of people to be selected if two particular people must be included = 5 – 2 = 3

Number of ways of selecting 3 out of (10 – 2) = 8 people = 8 C 3 = 8!/(3! × 5!) = 56.

∴ there are 56 ways of forming such a committee

Question 7:

How many triangles can be formed using 10 points in a plane out of which 4 are collinear?

Given 10 points on a plane out of which 4 are collinear, then 6 points are non-collinear.

Number of triangles formed by using 6 non-collinear points = 6 C 3 = 20

Number of triangles formed by using 6 non-collinear points and one out of the 4 collinear points = 6 C 2 × 4 C 1 = 15 × 4 = 60

Number of triangles formed by using 6 non-collinear points and two out of the 4 collinear points = 6 C 1 × 4 C 2 = 6 × 6 = 36

Total number of triangles can be formed = 20 + 60 + 36 = 116.

Question 8:

There are 4 balls of colour red, green, yellow and blue. In how many ways 2 two balls can be selected such that one of them is red or blue?

Number of ways two balls can be selected out of 4 balls = 4 C 2 = 6 ways

Number of ways two balls can be selected such that neither red nor blue ball gets selected = (4 – 2) C 2 = 2 C 2 = 1

Number of ways two balls can be selected such that either a red or a blue ball gets selected = 6 – 1 = 5 ways.

Question 9:

A team of four has to be selected from 6 boys and 4 girls. How many different ways a team can be selected if at least one boy must be there in the team?

Combination of a four-member team with at least one boy are:

{(BGGG), (BBGG), (BBBG), (BBBB)}

Number of ways one boy and three girls can be selected = 6 C 1 × 4 C 3 = 6 × 4 = 24

Number of ways two boys and two girls can be selected = 6 C 2 × 4 C 2 = 15 × 6 = 90

Number of ways three boys and one girl can be selected = 6 C 3 × 4 C 1 = 20 × 4 = 80

Number of ways four boys can be selected = 6 C 4 = 15

Total number of ways to form such a team = 24 + 90 + 80 + 15 = 209.

Question 10:

It is compulsory to answer 10 questions in an examination choosing at least 4 questions from each part A and part B. If there are 6 questions in part A and 7 questions in part B, in how many ways can 10 questions be attempted?

Case I: 4 questions from part A and 6 questions from part B

Number of ways of choosing 4 questions from part A = 6 C 4 = 15

Number of ways of choosing 6 questions from part B = 7 C 6 = 7

Total number of ways = 15 × 7 = 105

Case II: 5 questions from part A and 5 questions from part B

Number of ways of choosing 5 questions from part A = 6 C 5 = 6

Number of ways of choosing 5 questions from part B = 7 C 5 = 21

Total number of ways = 6 × 21 = 126

Case II: 6 questions from part A and 4 questions from part B

Number of ways of choosing 6 questions from part A = 6 C 6 = 1

Number of ways of choosing 4 questions from part B = 7 C 4 = 35

Total number of ways = 1 × 35 = 35

Required number of ways = 105 + 126 + 35 = 266.

## Practice Questions on Combinations

1. If 2n C r = 2n C r + 2 , find the value of r.

2. Prove that n. (n – 1) C (r – 1) = (n – r + 1). n C (r – 1) for all 1≤ r ≤ n.

3. How many diagonals are there of a 15-sided polygon?

4. How many ways a 6-member committee can be formed out of 12 people if two particular people must not be included?

5. A committee of 5 persons have to be formed out of 3 women and 6 men, such that there should be at most 3 women. How many ways can such a committee be formed?

6. How many sides are there of a convex polygon which have 44 diagonals?

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Quadratic equation, miscellaneous, permutations and combinations.

Permutation and Combination are the most fundamental concepts in mathematics and with these concepts, a new branch of mathematics is introduced to students i.e., combinatorics. Permutation and Combination are the ways to arrange a group of objects by selecting them in a specific order and forming their subsets. To arrange groups of data in a specific order permutation and combination formulas are used. Selecting the data or objects from a certain group is said to be permutation, whereas the order in which they are arranged is called a combination.

In this article we will study the concept of Permutation and Combination and their formulas, using these to solve many sample problems as well.

## Permutation

Permutation is the distinct interpretations of a provided number of components carried one by one, or some, or all at a time. For example, if we have two components A and B, then there are two likely performances, AB and BA.

A numeral of permutations when ‘r’ components are positioned out of a total of ‘n’ components is n P r . For example, let n = 3 (A, B, and C) and r = 2 (All permutations of size 2). Then there are 3 P 2 such permutations, which is equal to 6. These six permutations are AB, AC, BA, BC, CA, and CB. The six permutations of A, B, and C taken three at a time are shown in the image added below:

## Permutation Formula

Permutation formula is used to find the number of ways to pick r things out of n different things in a specific order and replacement is not allowed and is given as follows:

## Explanation of Permutation Formula

As we know, permutation is a arrengement of r things out of n where order of arrengement is important( AB and BA are two different permutation). If there are three different numerals 1, 2 and 3 and if someone is curious to permute the numerals taking 2 at a moment, it shows (1, 2), (1, 3), (2, 1), (2, 3), (3, 1), and (3, 2). That is it can be accomplished in 6 methods. Here, (1, 2) and (2, 1) are distinct. Again, if these 3 numerals shall be put handling all at a time, then the interpretations will be (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2) and (3, 2, 1) i.e. in 6 ways. In general, n distinct things can be set taking r (r < n) at a time in n(n – 1)(n – 2)…(n – r + 1) ways. In fact, the first thing can be any of the n things. Now, after choosing the first thing, the second thing will be any of the remaining n – 1 things. Likewise, the third thing can be any of the remaining n – 2 things. Alike, the r th thing can be any of the remaining n – (r – 1) things. Hence, the entire number of permutations of n distinct things carrying r at a time is n(n – 1)(n – 2)…[n – (r – 1)] which is written as n P r . Or, in other words,

## Combination

It is the distinct sections of a shared number of components carried one by one, or some, or all at a time. For example, if there are two components A and B, then there is only one way to select two things, select both of them.

For example, let n = 3 (A, B, and C) and r = 2 (All combinations of size 2). Then there are 3 C 2 such combinations, which is equal to 3. These three combinations are AB, AC, and BC.

Here, the combination of any two letters out of three letters A, B, and C is shown below, we notice that in combination the order in which A and B are taken is not important as AB and BA represent the same combination.

Note: In the same example, we have distinct points for permutation and combination. For, AB and BA are two distinct items i.e., two distinct permutation, but for selecting, AB and BA are the same i.e., same combination.

## Combination Formula

Combination Formula is used to choose ‘r’ components out of a total number of ‘n’ components, and is given by:

Using the above formula for r and (n-r), we get the same result. Thus,

## Explanation of Combination Formula

Combination, on the further hand, is a type of pack. Again, out of those three numbers 1, 2, and 3 if sets are created with two numbers, then the combinations are (1, 2), (1, 3), and (2, 3). Here, (1, 2) and (2, 1) are identical, unlike permutations where they are distinct. This is written as 3 C 2 . In general, the number of combinations of n distinct things taken r at a time is,

## Derivation of Permutation and Combination Formulas

We can derive these Permutation and Combination formulas using the basic counting methods as these formulas represent the same thing. Derivation of these formulas is as follows:

## Derivation of Permutations Formula

Permutation is selecting r distinct objects from n objects without replacement and where the order of selection is important, by the fundamental theorem of counting and the definition of permutation, we get

P (n, r) = n . (n-1) . (n-2) . (n-3). . . . .(n-(r+1))

By Multiplying and Dividing above with (n-r)! = (n-r).(n-r-1).(n-r-2). . . . .3. 2. 1, we get

P (n, r) = [n.(n−1).(n−2)….(nr+1)[(n−r)(n−r−1)(n-r)!] / (n-r)! ⇒ P (n, r) = n!/(n−r)!

Thus, the formula for P (n, r) is derived.

## Derivation of Combinations Formula

Combination is choosing r items out of n items when the order of selection is of no importance. Its formula is calculated as,

C(n, r) = Total Number of Permutations /Number of ways to arrange r different objects. [Since by the fundamental theorem of counting, we know that number of ways to arrange r different objects in r ways = r!] C(n,r) = P (n, r)/ r! ⇒ C(n,r) = n!/(n−r)!r!

Thus, the formula for Combination i.e., C(n, r) is derived.

## Difference Between Permutation and Combination

Various differences between the permutation and combination can be understood by the following table:

## Solved Examples on Permutation and Combination

Example 1: Find the number of permutations and combinations of n = 9 and r = 3 .

Solution:

Given, n = 9, r = 3 Using the formula given above: For Permutation: n P r = (n!) / (n – r)! ⇒ n P r = (9!) / (9 – 3)! ⇒ n P r = 9! / 6! = (9 × 8 × 7 × 6! )/ 6! ⇒ n P r = 504 For Combination: n C r = n!/r!(n − r)! ⇒ n C r = 9!/3!(9 − 3)! ⇒ n C r = 9!/3!(6)! ⇒ n C r = 9 × 8 × 7 × 6!/3!(6)! ⇒ n C r = 84

Example 2: In how many ways a committee consisting of 4 men and 2 women, can be chosen from 6 men and 5 women?

Choose 4 men out of 6 men = 6 C 4 ways = 15 ways Choose 2 women out of 5 women = 5 C 2 ways = 10 ways The committee can be chosen in 6 C 4 × 5 C 2 = 150 ways.

Example 3: In how many ways can 5 different books be arranged on a shelf?

This is a permutation problem because the order of the books matters. Using the permutation formula, we get: 5 P 5 = 5! / (5 – 5)! = 5! / 0! = 5 x 4 x 3 x 2 x 1 = 120 Therefore, there are 120 ways to arrange 5 different books on a shelf.

Example 4: How many 3-letter words can be formed using the letters from the word “FABLE”?

This is a permutation problem because the order of the letters matters. Using the permutation formula, we get: 5 P 3 = 5! / (5 – 3)! = 5! / 2! = 5 x 4 x 3 = 60 Therefore, there are 60 3-letter words that can be formed using the letters from the word “FABLE”.

Example 5: A committee of 5 members is to be formed from a group of 10 people. In how many ways can this be done?

This is a combination problem because the order of the members doesn’t matter. Using the combination formula, we get: 10 C 5 = 10! / (5! x (10 – 5)!) = 10! / (5! x 5!) ⇒ 10 C 5 = (10 x 9 x 8 x 7 x 6) / (5 x 4 x 3 x 2 x 1) = 252 Therefore, there are 252 ways to form a committee of 5 members from a group of 10 people.

Example 6: A pizza restaurant offers 4 different toppings for their pizzas. If a customer wants to order a pizza with exactly 2 toppings, in how many ways can this be done?

This is a combination problem because the order of the toppings doesn’t matter. Using the combination formula, we get: 4 C 2 = 4! / (2! x (4 – 2)!) = 4! / (2! x 2!) = (4 x 3) / (2 x 1) = 6 Therefore, there are 6 ways to order a pizza with exactly 2 toppings from 4 different toppings.

Example 7: How considerable words can be created by using 2 letters from the term“LOVE”?

The term “LOVE” has 4 distinct letters. Therefore, required number of words = 4 P 2 = 4! / (4 – 2)! Required number of words = 4! / 2! = 24 / 2 ⇒ Required number of words = 12

Example 8: Out of 5 consonants and 3 vowels, how many words of 3 consonants and 2 vowels can be formed?

Number of ways of choosing 3 consonants from 5 = 5 C 3 Number of ways of choosing 2 vowels from 3 = 3 C 2 Number of ways of choosing 3 consonants from 2 and 2 vowels from 3 = 5 C 3 × 3 C 2 ⇒ Required number = 10 × 3 = 30 It means we can have 30 groups where each group contains a total of 5 letters (3 consonants and 2 vowels). Number of ways of arranging 5 letters among themselves = 5! = 5 × 4 × 3 × 2 × 1 = 120 Hence, the required number of ways = 30 × 120 ⇒ Required number of ways = 3600

Example 9: How many different combinations do you get if you have 5 items and choose 4?

Insert the given numbers into the combinations equation and solve. “n” is the number of items that are in the set (5 in this example); “r” is the number of items you’re choosing (4 in this example): C(n, r) = n! / r! (n – r)! ⇒ n C r = 5! / 4! (5 – 4)! ⇒ n C r = (5 × 4 × 3 × 2 × 1) / (4 × 3 × 2 × 1 × 1) ⇒ n C r = 120/24 ⇒ n C r = 5 The solution is 5.

Example 10: Out of 6 consonants and 3 vowels, how many expressions of 2 consonants and 1 vowel can be created?

Number of ways of selecting 2 consonants from 6 = 6 C 2 Number of ways of selecting 1 vowels from 3 = 3 C 1 Number of ways of selecting 3 consonants from 7 and 2 vowels from 4. ⇒ Required ways = 6 C 2 × 3 C 1 ⇒ Required ways = 15 × 3 ⇒ Required ways= 45 It means we can have 45 groups where each group contains a total of 3 letters (2 consonants and 1 vowels). Number of ways of arranging 3 letters among themselves = 3! = 3 × 2 × 1 ⇒ Required ways to arrenge three letters = 6 Hence, the required number of ways = 45 × 6 ⇒ Required ways = 270

Example 11: In how many distinct forms can the letters of the term ‘PHONE’ be organized so that the vowels consistently come jointly?

The word ‘PHONE’ has 5 letters. It has the vowels ‘O’,’ E’, in it and these 2 vowels should consistently come jointly. Thus these two vowels can be grouped and viewed as a single letter. That is, PHN(OE). Therefore we can take total letters like 4 and all these letters are distinct. Number of methods to organize these letters = 4! = 4 × 3 × 2 × 1 ⇒ Required ways arrenge letters = 24 All the 2 vowels (OE) are distinct. Number of ways to arrange these vowels among themselves = 2! = 2 × 1 ⇒ Required ways to arrange vowels = 2 Hence, the required number of ways = 24 × 2 ⇒ Required ways = 48.

## FAQs on Permutations and Combinations

Q1: what is the factorial formula.

Factorial formula is used for the calculation of permutations and combinations. The factorial formula for n! is given as n! = n × (n-1) × . . . × 4 × 3 × 2 × 1 For example, 3! = 3 × 2 × 1 = 6 and 5! = 5 × 4 × 3 × 2 × 1 = 120.

## Q2: What does n C r represent?

n C r represents the number of combinations that can be made from “n” objects taking “r” at a time.

## Q3: What do you mean by permutations and combinations?

A permutation is an act of arranging things in a specific order. Combinations are the ways of selecting r objects from a group of n objects, where the order of the object chosen does not affect the total combination.

## Q4: Write examples of permutations and combinations.

The number of 3-letter words that can be formed by using the letters of the word says, HELLO; 5 P 3 = 5!/(5-3)! this is an example of a permutation. The number of combinations we can write the words using the vowels of the word HELLO; 5 C 2 =5!/[2! (5-2)!], this is an example of a combination.

## Q5: Write the formula for finding permutations and combinations.

Formula for calculating permutations: n Pr = n!/(n-r)! Formula for calculating combinations: n Cr = n!/[r! (n-r)!]

## Q6: Write some real-life examples of permutations and combinations.

Sorting of people, numbers, letters, and colors are some examples of permutations. Selecting the menu, clothes, and subjects, are examples of combinations.

## Q7: What is the value of 0!?

The value of 0! = 1, is very useful in solving the permutation and combination problems.

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## COMMENTS

Solve the equation to find the number of combinations. You can do this either by hand or with a calculator. If you have a calculator available, find the factorial setting and use that to calculate the number of combinations. If you're using Google Calculator, click on the x! button each time after entering the necessary digits.

Combinations. There are also two types of combinations (remember the order does not matter now): Repetition is Allowed: such as coins in your pocket (5,5,5,10,10) No Repetition: such as lottery numbers (2,14,15,27,30,33) 1. Combinations with Repetition. Actually, these are the hardest to explain, so we will come back to this later. 2.

10,000 combinations. First method: If you count from 0001 to 9999, that's 9999 numbers. Then you add 0000, which makes it 10,000. Second method: 4 digits means each digit can contain 0-9 (10 combinations). The first digit has 10 combinations, the second 10, the third 10, the fourth 10. So 10*10*10*10=10,000.

Examples of solving Combination Problems with videos and solutions, Formula to find the number of combinations of n things taken r at a time, What is the Combination Formula, How to use the Combination Formula to solve word problems and counting problems, How to solve combination problems that involve selecting groups based on conditional criteria, How to solve word problems involving ...

Probability & combinations (2 of 2) Example: Different ways to pick officers. Example: Combinatorics and probability. Getting exactly two heads (combinatorics) Exactly three heads in five flips. Generalizing with binomial coefficients (bit advanced) Example: Lottery probability. Conditional probability and combinations.

total outcome= 2^5=32 (since every throw might be basket or a miss, 2 possibility for every throw). combination of choosing 3 out of 5= 5!/3!2!= 10. total probability = 10/32=31.25% but the answer is 20.48%....does it have to do something odds of scoring a basket or missing is not equal. •.

Combinations refer to the possible arrangements of a set of given objects when changing the order of selection of the objects is not treated as a distinct arrangement. After reading this article, you should understand: Combination Formula and its derivation; Difference between permutation and combination; How to solve problems related to ...

Note: The difference between a combination and a permutation is whether order matters or not. If the order of the items is important, use a permutation. If the order of the items is not important, use a combination. Now here are a couple examples where we have to figure out whether it is a permuation or a combination.

Learn how to solve mathematics word problems containing Permutations and Combinations using formulas. ... Step by step guide to solve Permutations and Combinations. Permutations: The number of ways to choose a sample of \(k\) elements from a set of \(n\) distinct objects where order does matter, and replacements are not allowed. ...

Find the number of strings of 4 letters that can be formed with the letters of the word EXAMINATION? Solution : There are 11 letters not all different. They are AA, II, NN, E, X, M, T, O. The following combinations are possible: Case 1 : Number of ways selecting 2 alike, 2 alike. = 3C2 = 3 ways. Case 2 :

Learn how to solve problems involving permutations and combinations in this tutorial video. You will see examples of how to apply the formulas and rules for counting different arrangements and ...

Example Question From Combination Formula. Question 1: Father asks his son to choose 4 items from the table. If the table has 18 items to choose, how many different answers could the son give? Therefore, simply: find "18 Choose 4". We know that, Combination = C (n, r) = n!/r! (n-r)!

Example 7.5.2 7.5. 2 Example with Restrictions. From an unlimited selection of five types of soda, one of which is Dr. Pepper, you are putting 25 cans on a table. (a) Determine the number of ways you can select 25 cans of soda. Solution. (b) Determine the number of ways you can select 25 cans of soda if you must include at least seven Dr. Peppers.

Formula for permutation: nPr = n!/(n-r)! Formula for combination: nCr = n!/r!.(n-r)! Difference between permutation and combination *In permutations, the order matters*, so rearranging the order of selected objects results in different permutations.*In combinations, the order does not matter*, so different arrangements of the same set of objects are considered equivalent.

Learn how to recognize when to use the combination formula. Permutations and combinations are similar yet different. When using a combination the order is no...

This is a combination problem: combining 2 items out of 3 and is written as follows: nC r = n! (n − r)! r! n C r = n! ( n − r)! r! The number of combinations is equal to the number of permutations divided by r! to eliminate those counted more than once because the order is not important. Example 7: Calculate. a) 3C 2 3 C 2.

Solving Word Problems Involving Combinations: Example 2 A committee of 5 members must be chosen from a track club. The club has 15 sprinters, 9 jumpers, and 7 long-distance runners.

Combination questions with solutions are given here to practice and to understand how and when to use the concept of combinations while solving a problem. Also, try important permutation and combination questions for class 11. In combinatorics, the combination is a way of selecting something from a given collection. For example, we have to form ...

To calculate the number of combinations with repetitions, use the following equation. Where: Advertisement. n = the number of options. r = the size of each combination. The exclamation mark (!) represents a factorial. In general, n! equals the product of all numbers up to n. For example, 3! = 3 * 2 * 1 = 6.

Using Permutation vs Combination to Solve Probability Problems. This overview just scratches the surface of using permutations and combinations. For more detail about each one along with worked examples for solving probability problems, please read my separate articles about: Using Permutations to Solve Probability Problems

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This is a combination problem because the order of the members doesn't matter. Using the combination formula, we get: 10 C 5 = 10! / ... Insert the given numbers into the combinations equation and solve. "n" is the number of items that are in the set (5 in this example); "r" is the number of items you're choosing (4 in this example

Handshake Problem as a Combinations Problem. We can also solve this Handshake Problem as a combinations problem as C(n,2). n (objects) = number of people in the group r (sample) = 2, the number of people involved in each different handshake The order of the items chosen in the subset does not matter so for a group of 3 it will count 1 with 2, 1 ...