how to solve problems with combination

• PRO Courses Guides New Tech Help Pro Expert Videos About wikiHow Pro Upgrade Sign In
• EDIT Edit this Article
• EXPLORE Tech Help Pro About Us Random Article Quizzes Request a New Article Community Dashboard This Or That Game Popular Categories Arts and Entertainment Artwork Books Movies Computers and Electronics Computers Phone Skills Technology Hacks Health Men's Health Mental Health Women's Health Relationships Dating Love Relationship Issues Hobbies and Crafts Crafts Drawing Games Education & Communication Communication Skills Personal Development Studying Personal Care and Style Fashion Hair Care Personal Hygiene Youth Personal Care School Stuff Dating All Categories Arts and Entertainment Finance and Business Home and Garden Relationship Quizzes Cars & Other Vehicles Food and Entertaining Personal Care and Style Sports and Fitness Computers and Electronics Health Pets and Animals Travel Education & Communication Hobbies and Crafts Philosophy and Religion Work World Family Life Holidays and Traditions Relationships Youth
• Browse Articles
• Learn Something New
• Quizzes Hot
• This Or That Game New
• Train Your Brain
• Explore More
• Support wikiHow
• About wikiHow
• Log in / Sign up
• Education and Communications
• Mathematics

How to Calculate Combinations

Last Updated: November 7, 2023 References

wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. To create this article, volunteer authors worked to edit and improve it over time. There are 7 references cited in this article, which can be found at the bottom of the page. This article has been viewed 121,716 times. Learn more...

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.

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/

About This Article

• Send fan mail to authors

Did this article help you?

Featured Articles

Trending Articles

Watch Articles

• Terms of Use
• Privacy Policy
• Do Not Sell or Share My Info
• Not Selling Info

Don’t miss out! Sign up for

wikiHow’s newsletter

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?

We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.

Mathematics

• Algebraic Expressions
• Linear Equations
• Parametric Equations
• Absolute-value Equations
• Sets of Equations
• Practical Problems
• Linear Inequations
• Linear Inequations - Tab
• Quadratic Equations
• Quadratic Equation by Discussion
• Quadratic Equation - Roots Properties
• Irrational Equations
• Quadratic Inequalities
• Absolute Value
• Exponential Equations
• Logarithms - Basics
• Logarithmic Exponential Equations
• Logarithmic Equations - Other Bases
• Quadratic Logarithmic Equations
• Sets of Logarithmic Equations
• Expressions
• Súčtové a rozdielové vzorce
• Dvojnásobný a polovičný argument
• Arithmetic Progression
• Geometric Progression
• Factorial & Binomial Coefficient
• Binomial Theorem
• Permutations (general)
• Permutations

Combinations

• Probability
• Right Triangle

1. Characterize combinations and combinations with repetition.

2. On the plane there are 6 different points (no 3 of them are lying on the same line). How many segments do you get by joining all the points?

3. On a circle there are 9 points selected. How many triangles with edges in these points exist?

4. a) Find out a formula for counting the number of diagonals in a convex n-gon! b) How many diagonals has a 10-gon?

5. In how many ways you can choose 8 of 32 playing cards not considering their order?

The playing cards can be chosen in 10 518 300 ways.

6. A teacher has prepared 20 arithmetics tasks and 30 geometry tasks. For a test he‘d like to use:

7. On a graduation party the graduants pinged their glasses. There were 253 pings. How many graduants came to the party?

8. If the number of elements would raise by 8, number of combinations with k=2 without repetition would raise 11 times. How many elements are there?

9. For which x positive integer stands:

10. Two groups consist of 26 elements and 160 combinations without repetition for k=2 together. How many elements are in the first and how many in the second group?

11. In the confectioners 5 different icecreams are sold. A father would like to buy 15 caps of icecream for his family. In how many ways can he buy the icecream?

12. From how many elements 15 combinations with repetition (k=2) can be made?

Adblock detected

Vážený návštevník Priklady.eu, našim systémom bolo detekované odmietnutie zobrazenie reklamy. Na vašom počítači je teda veľmi pravdepodobne nainštalovaný softvér slúžiaci na blokovanie reklám. Reklamy sú pre nás jediným zdrojom príjmov, čo nám umožňuje poskytovať Vám obsah bez poplatkov, zadarmo. Prosíme, odblokujte ho. Povolenie reklamy na tejto stránke je možné docieliť aktiváciou voľby "Nespúšťať Adblock na stránkach na tejto doméne", alebo "Vypnúť Adblock na priklady.eu", prípadne inú podobnú položkou v menu vášho programu na blokovanie reklám. Ďakujeme za pochopenie, tím Priklady.eu. Vážený návštěvníku Priklady.eu, našim systémem bylo detekováno odmítnutí zobrazení reklamy. Na vašem počítači je tedy velice pravděpodobně nainstalován software sloužící k blokování reklam. Reklamy jsou pro nás jediným zdrojem příjmů, což nám umožňuje Vám poskytovat obsah bez poplatků, zdarma. Prosíme, odblokujte je. Povolení reklamy na této stránce lze docílit aktivací volby "Nespouštět AdBlock na stránkách na této doméně", nebo "Vypnout AdBlock na priklady.eu", případně jinou podobnou položkou v menu vašeho programu na blokování reklam. Děkujeme za pochopení, tým Priklady.eu.

Combination Word Problems

Exercise 10, exercise 11, exercise 12, solution of exercise 1, solution of exercise 2, solution of exercise 3, solution of exercise 4, solution of exercise 5, solution of exercise 6, solution of exercise 7, solution of exercise 8, solution of exercise 9, solution of exercise 10, solution of exercise 11, solution of exercise 12.

How many different combinations of management can there be to fill the positions of president, vice-president and treasurer of a football club knowing that there are 12 eligible candidates?

How many different ways can the letters in the word "micro" be arranged if it always has to start with a vowel?

How many combinations can the seven colors of the rainbow be arranged into groups of three colors each?

How many different five-digit numbers can be formed with only odd numbered digits? How many of these numbers are greater than 70,000?

How many games will take place in a league consisting of four teams? (Each team plays each other twice, once at each teams respective "home" location)

10 people exchange greetings at a business meeting. How many greetings are exchanged if everyone greets each other once?

How many five-digit numbers can be formed with the digits 1, 2 and 3? How many of those numbers are even?

How many lottery tickets must be purchased to complete all possible combinations of six numbers, each with a possibility of being from 1 to 49?

How many ways can 11 players be positioned on a soccer team considering that the goalie cannot hold another position other than in goal?

How many groups can be made from the word "house" if each group consists of 3 alphabets?

Sarah has 8 colored pencils that are all unique. She wants to pick three colored pencils from her collection and give them to her younger sister. How many different combinations of colored pencils can Sarah make from 8 pencils?

Alice has 6 chocolates. All of the chocolates are of different flavors. She wants to give two of her chocolates to her friend. How many different combinations of chocolates can Alice make from six chocolates?

The order of the elements does matter.

The elements cannot be repeated.

The words will begin with i or o followed by the remaining 4 letters taken from 4 by 4.

The order of the elements does not matter.

n = 5      k = 5

The odd numbers greater than 70,000 have to begin with 7 or 9. Therefore:

The elements are repeated.

If the number is even it can only end in 2.

Therefore, there are 10 players who can occupy 10 different positions.

The word house has 5 alphabets. If each new word should have 3 alphabets, then we should use the following formula:

Substitute the values in this example in the above formula:

Number of pencils Sarah have = 8

Number of pencils she wants to give to her younger sister = 3

We will use the binomial coefficients formula to determine the number of combinations:

After substitution we will get the number of combinations:

Hence, Sarah can make 56 combinations of 8 colored pencils given the fact that she can choose 3 at a time.

Number of chocolates Alice have = 6

Number of chocolates she wants to give to her friend= 2

Hence, Alice can make 15 combinations of 6 chocolates given the fact that she can choose 2.

Did you like this article? Rate it!

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.

Combinations

Combinatorics, permutations, circular permutations, permutations with repetition, combinations with repetition, pascal’s triangle, combinatorics formulas, permutations word problems, cancel reply.

Your comment

Current ye@r *

Leave this field empty

I need to learn

I must say everything is superb I wish I could access to most of my maths topics here! Questions are easy to comprehend. Thanks for your help.

in how many ways can be 5 red balls and green balls be arranged in a row of no two green balls can be next to each other?

Great job but explanations are too complex

• Pre-algebra lessons
• Pre-algebra word problems
• Algebra lessons
• Algebra word problems
• Algebra proofs
• Advanced algebra
• Geometry lessons
• Geometry word problems
• Geometry proofs
• Trigonometry lessons
• Consumer math
• Baseball math
• Math for nurses
• Statistics made easy
• High school physics
• Basic mathematics store
• SAT Math Prep
• Math skills by grade level
• Ask an expert
• Other websites
• K-12 worksheets
• Worksheets generator
• Algebra worksheets
• Geometry worksheets
• Free math problem solver
• Pre-algebra calculators
• Algebra Calculators
• Geometry Calculators
• Math puzzles
• Member login

Combination word problems

Here are some carefully chosen combination word problems that will show you how to solve word problems involving combinations.  

Use the combination formula shown below when the order does not matter

The combination word problems will show you how to do the followings:

• Use the combination formula
• Use the multiplication principle and the combination formula
• Use the addition principle and the combination formula
• Use the multiplication principle, addition principle, and combination formula

Word problem #1 There are 18 students in a classroom. How many different eleven-person students can be chosen to play in a soccer team?

Solution The order in which students are listed once the students are chosen does not distinguish one student from another. You need the number of combinations of 18 potential students chosen 11 at a time. Evaluate n C r with n = 18 and r = 11

18 C 11 = 31824

There are 31824 different eleven-person students that can be chosen from a group of 18 students.

Word problem #2 For your biology report, you can choose to write about three of a list of four different animals. Find the number of combinations possible for your report.

Solution The order in which you write about these 3 animals does not matter as long as you write about 3 animals. Evaluate  n C r  with n = 4 and r = 3

There are 4 different ways you can choose 3 animals from a list of 4.

Word problem #3 A math teacher would like to test the usefulness of a new math game on 4 of the 10 students in the classroom. How many different ways can the teacher pick students? 

Solution The order in which the teacher picks students does not matter.

Evaluate  n C r  with n = 10 and r = 4

There are 210 ways the teacher can pick students

More challenging combination word problems

These combination word problems will also show you how to use the multiplication principle and the addition principle.

Word problem #4 A company has 20 male employees and 30 female employees. A grievance committee is to be established. If the committee will have 3 male employees and 2 female employees, how many ways can the committee be chosen?

This problem has the following two tasks:

Task 1 : choose 3 males from 20 male employees

Task 2: choose 2 females from 30 female employees

We need to use the fundamental counting principle , also called the multiplication principle, since we have more than 1 task.  

Fundamental counting principle

If you have n choices for a first task and m choices for a second task, you have n × m choices for both tasks.

Therefore, evaluate 20 C 3 and 30 C 2 and then multiply  20 C 3 by 30 C 2

20 C 3  ×  30 C 2 = 1140 × 435 = 495900

The number of ways the committee can be chosen is 495900

Word problem #5 Eight candidates are competing to get a job at a prestigious company.  The company has the freedom to choose as many as two candidates. In how many ways can the company choose two or fewer candidates.  Solution

The company can choose 2 people, 1 person, or none.

Notice that this time we need to use the addition principle as opposed to using the multiplication principle.

What is the difference? The key difference here is that the company will choose either 2, 1, or none. The company will not choose 2 people and 1 person at the same time. This does not make sense!

Addition principle

Let A and B be two events that cannot happen together. If n is the number of choices for A and m is the number of choices for B, then n + m is the number of choices for A and B. 

Therefore you need to evaluate  8 C 2 ,  8 C 1 , and  8 C 0 and then add  8 C 2 ,  8 C 1 , and  8 C 0 together.

Useful shortcuts to find combinations

n C 1 = n and n C 0 = 1

Therefore,  10 C 1  = 10 and  10 C 0  = 1 8 C 2 + 8 C 1 + 8 C 0 = 28 + 10 + 1 = 39 

The company has 39 ways to choose two or fewer candidates. 

Word problem #6 A company has 20 male employees and 30 female employees. A grievance committee is to be established. If the committee will have as many as 3 male employees and as many as 2 female employees, how many ways can the committee be chosen?

The expression as many as  makes the problem quite complex now since we now have all the following cases to consider.

Choose 3 males, 2 males, 1 male, or 0 male

Choose 2 females, 1 female, or 0 female.

Here is a complete list of all the different cases.

• 3 males and 2 females
• 3 males and 1 female
• 3 males and 0 female
• 2 males and 2 females
• 2 males and 1 female
• 2 males and 0 female
• 1 male and 2 females
• 1 male and 1 female
• 1 male and 0 female
• 0 male and 2 females
• 0 male and 1 female
• 0 male and 0 female

We only need to find  20 C 2

3 males and 2 females:  20 C 3  ×  30 C 2  = 1140 × 435 = 495900 (done in problem #4 )

3 males and 1 female:  20 C 3  ×  30 C 1 = 1140 × 30 = 34200 

3 males and 0 female:  20 C 3  ×  30 C 0 = 1140 × 1 = 1140 

2 males and 2 females:  20 C 2 ×  30 C 2 =  190 × 435 = 82650

2 males and 1 female:  20 C 2  ×  30 C 1 = 190 × 30 = 5700

2 males and 0 female:  20 C 2 ×  30 C 0 = 190 × 1 = 190 

1 male and 2 females:  20 C 1 ×  30 C 2 = 20 × 435 = 8700 

1 male and 1 female:  20 C 1  ×  30 C 1 = 20 × 30 = 600 

1 male and 0 female:  20 C 1  ×  30 C 0  = 20 × 1 = 20 

0 male and 2 females:  20 C 0 ×  30 C 2 = 1 × 435 = 435

0 male and 1 female:  20 C 0  ×  30 C 1 = 1 × 30 = 30

0 male and 0 female: 20 C 0 × 30 C 0 = 1 × 1 = 1

Add everything:

495900 + 34200 + 1140 + 82650 + 5700 + 190 + 8700 + 600 + 20 + 435 + 30 + 1 = 629566.

The number of ways to choose the committee is 629566

How to find the number of combinations

Recent Articles

Percent of increase word problems.

Oct 27, 23 07:50 AM

What is Algebra? Definition and Examples

Oct 11, 23 02:49 PM

How To Find The Factors Of 20: A Simple Way

Sep 17, 23 09:46 AM

100 Tough Algebra Word Problems. If you can solve these problems with no help, you must be a genius!

 Recommended

About me :: Privacy policy :: Disclaimer :: Donate   Careers in mathematics  

Copyright © 2008-2021. Basic-mathematics.com. All right reserved

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":