# Combination: Definition, Formula & Examples

An error occurred trying to load this video.

Try refreshing the page, or contact customer support.

Coming up next: Permutation & Combination: Problems & Practice

### You're on a roll. Keep up the good work!

Replay
Your next lesson will play in 10 seconds
• 0:00 Combinations and Permutations
• 1:11 Factorial Notation
• 1:47 Combination Formulas
• 2:52 Solving Combination Problems
• 5:42 Lesson Summary
Save Save

Want to watch this again later?

Timeline
Autoplay
Autoplay
Speed Speed Audio mode

#### Recommended Lessons and Courses for You

Lesson Transcript
Instructor
Laura Pennington

Laura received her Master's degree in Pure Mathematics from Michigan State University. She has 15 years of experience teaching collegiate mathematics at various institutions.

Expert Contributor
Kathryn Boddie

Kathryn earned her Ph.D. in Mathematics from UW-Milwaukee in 2019. She has over 10 years of teaching experience at high school and university level.

In this lesson, we'll learn how to identify combinations of objects. We'll also look at various formulas that allow us to calculate the number of possible combinations in a given scenario.

## Combinations and Permutations

In mathematics, combinations and permutations are normally studied at the same time because they are very similar. For instance, both permutations and combinations are collections of objects. But while a combination is a collection of the objects where the order doesn't matter, a permutation is an arrangement of a group of objects where the order does matter.

For example, suppose you were to choose four from a group of 20 people: Andrea, Alex, Sophie, and Nathan. If we said you chose Alex, Nathan, Sophie and Andrea, we'd still be talking about the same group of people. As the order in which we name them wouldn't matter, we'd be referring to a combination of four people.

Now, imagine that your employer just assigned you the following 4-digit employee identification code: 4793. If we changed the order of the digits to 9734, we'd end up with a different ID number. As the order of the digits does matter, 4793 is a permutation of four digits.

## Factorial Notation

Before we get into the combination formulas, we first need to talk about n!, not in an excited fashion, which you may have assumed from the exclamation point, but in a mathematical context. In math, n!, pronounced n factorial, represents the product of all of the integers from n down to 1, as shown in the image below:

For example, 4! = 4 * 3 * 2 * 1 = 24. We'll be using n! a lot when dealing with permutations and combinations.

## Combination Formulas

When it comes to combination formulas, there are two scenarios we want to consider:

1. Repetition is allowed
2. Repetition is not allowed

To understand what these scenarios mean, let's revisit the group of people we discussed at the beginning of the lesson. Imagine that Alex was the first of the four we chose to be in our group. Once we've chosen Alex, we can't choose him again since he's already in the group. So, in this scenario, repetition is not allowed.

Now, pretend you're at a sushi bar that offers 11 different types of sushi, from which you can choose three. Suppose the first type of sushi you choose is salmon. As the menu allows you to have 3 pieces of sushi, your second and third choices could also be salmon. In this scenario, repetition is allowed.

It is important to understand whether or not repetition is allowed when determining which formula to use when solving combination problems. The formulas we use when dealing with combinations are described in the image below:

## Solving Combination Problems

To solve problems involving combinations, we follow these steps:

1. Make sure you are dealing with a combination problem, where order does not matter, and not a permutation problem
2. Determine if repetition is allowed
3. Use the appropriate formula based on what you found in the second step
4. Substitute known numbers for the values in the formula, and perform the operations

To unlock this lesson you must be a Study.com Member.

## Preliminary Discussion

There are 52 playing cards in a standard deck of cards. Of the 52 cards, there are 13 hearts, 13 spades, 13 clubs, and 13 diamonds. The hearts and diamonds are red, and the clubs and spades are black. Each suit has cards 2 through 10, A, J, Q, K.

Since 25% of the playing cards are hearts, your friend thinks that the number of 5 card hands that are all hearts will be 25% of the total number of 5 card hands. This same friend says that since half the cards are red, the number of 5 card hands made up of only red cards is half of the total number of 5 card hands. Without calculating the number of hands, discuss whether you think your friend is correct.

### Math Problems to Answer the Discussion

Work through the problems below to discover whether your thoughts from the above discussion were correct.

1. If you have a standard deck of 52 cards, how many different 5 card hands can be dealt?

2. There are 13 hearts. How many different hands of 5 cards can be made only of hearts?

3. Half of the cards are red. How many different hands of 5 cards can be made with only red cards?

4. Use the answers from 1. and 2. to find the percentage of total card hands made up by hands of only hearts.

5. Use the answers from 1. and 3. to find the percentage of total card hands made up by hands of only red cards.

## Solutions

1. The order of the cards in a hand does not matter but we cannot repeat cards in the same hand, so we need to find the number of combinations without repetition for {eq}n=52 {/eq} and {eq}r=5 {/eq} by evaluating {eq}\dfrac{n!}{r!(n-r)!}=\dfrac{52!}{5!(52-5)!} = \dfrac{52!}{5!47!} = \dfrac{52\times 51\times 50\times 49\times 48\times 47!}{5!47!} = \dfrac{52\times 51\times 50\times 49\times 48}{5\times 4\times 3\times 2\times 1} = \dfrac{311875200}{120} = 2598960 {/eq}

There are 2,598,960 different 5 card hands.

2. We need to find {eq}\dfrac{n!}{r!(n-r)!} {/eq} with {eq}n=13, r=5 {/eq}

{eq}\dfrac{n!}{r!(n-r)!}=\dfrac{13!}{5!(13-5)!} = \dfrac{13!}{5!8!} = \dfrac{13\times 12\times 11\times 10\times 9}{5\times 4\times 3\times 2\times 1} =\dfrac{154440}{120}=1287 {/eq}

There are 1,287 different 5 card hands made only of hearts.

3. We need to find {eq}\dfrac{n!}{r!(n-r)!} {/eq} with {eq}n=26, r=5 {/eq}

{eq}\dfrac{n!}{r!(n-r)!}=\dfrac{26!}{5!(26-5)!} = \dfrac{26!}{5!21!} = \dfrac{26\times 25\times 24\times 23\times 22}{5\times 4\times 3\times 2\times 1} =\dfrac{7893600}{120}=65780 {/eq}

There are 65,780 different 5 card hands made only of red cards.

4. The percentage of 5 card hands made only of hearts is given by {eq}\dfrac{1287}{2598960}\approx 0.000495 {/eq}

About 0.0495% of all 5 card hands are made only of hearts.

5. The percentage of 5 card hands made only of red cards is given by {eq}\dfrac{65780}{2598960}\approx 0.0253 {/eq}

About 2.53% of all 5 card hands are made only of red cards.

Looking at the above results, we can conclude that your friend is incorrect. Your friend thought that the percentage of 5 card hands containing only hearts would be 25%, but it is actually only 0.0495% and your friend thought that the percentage of 5 card hands made of only red cards would be 50%, but it is actually only 2.53%

### Register to view this lesson

Are you a student or a teacher?

#### See for yourself why 30 million people use Study.com

##### Become a Study.com member and start learning now.
Back
What teachers are saying about Study.com

### Earning College Credit

Did you know… We have over 200 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level.