Do you know the difference between permutation and combination? No? You’re not alone. Combinations and permutations are the bane of many students. Yet, what I’ve noticed over the years is it’s not so much both of them that are the issue as it is which one to use for a particular problem: the combination vs permutation question. In other words, students have no difficult identifying whether a question is a combinations/permutations problem. The difficulty is in knowing exactly which one it is—combinations or permutations?

## Combinations and Permutations

One way to think of it is to think of **permutations** as the number of arrangements or orderings within a fixed group. For example, if I have five students and I want to figure out how many ways they can sit in five chairs, I’m going to use the permutations formula. First off, the number in the group is fixed. Secondly, I’m looking for how many ways I can “arrange” the students in five chairs.

**Combinations**, on the other hand, are useful when figuring out how many groups I can form from a larger number of people. For instance, if I’m a basketball coach and I want to find out how many distinct teams I can form based on a group of people, I want to use combinations.

To make sure you understand this important distinction, here are three different scenarios. Your job is not to solve the question but to determine whether you use the combinations or the permutations formula to solve them.

## Combination vs Permutation: You Decide

1. Joan has five panels at home that she wants to paint. She has five different colored paints and intends to paint each panel a different color. How many different ways can she paint the five panels?

2. How many unique combinations of the word MAGOOSH can I form by scrambling the letters?

3. There are seven astronauts who are trying out to be part of three-person in-space flight team. How many different flight teams can be formed?

## Difference Between Combination and Permutation: Answers

1. Permutations

She wants to arrange the colors. The number of panels is fixed. Had she been choosing five panels from a total of 8, let’s say, then we would need to use combinations.

2. Permutations

Okay, this was a little bit of a trick, since I used the word “combinations”. But that word I used colloquially, not mathematically. In this case, the number of letters is fixed. We are simply rearranging them.

3. Combinations

We are choosing a group from a larger group. One way to think of it is that when you use the word choose in the context of selecting from a group, you are dealing with combinations. And “choose” and “combinations” both begin with the letter ‘C’. There is an exception to this rule that I’ll talk about in the next section.

## Difference Between Permutation and Combination: The Main Point

Big Idea: If you are forming a group from a larger group and the placement within the smaller group is important, then you want to use permutations.

Imagine a group of 12 sprinters is competing for the gold medal. During the award ceremony, a gold medal, silver medal and bronze medal will be awarded. How many different ways can these three medals be handed out?

Remember that with permutations ordering is key. Even though the top three spots for the sprinters forms a subgroup, it is the ordering within that subgroup that matters greatly, and is the difference between a gold, silver, and a bronze medal. An easy way to solve this question mathematically is to imagine that the dashes below are the podium upon which each sprinter will stand (albeit the dashes are at the same level):

____ ____ ____

gold silver bronze

To find out the number of different arrangement, ask yourself how many athletes can stand on the gold medal podium? Well, we have a total of 12 athletes. What about the silver medal podium? Now, we have one athlete fewer—since one is already on the gold medal podium. So that gives us a total of 11 for the silver medal spot. Finally, that leaves us with 10 athletes for the bronze medal podium.

The math looks like this:

12 x 11 x 10 = 1320

You might notice that this is the fundamental counting principle. The idea that when we are looking for total number of outcomes, we multiply numbers—or in this case, the number that goes on top of each dash—together. For instance, if I have six pairs of shorts and four pairs of t-shirts and I’m wondering how many different combinations of shorts and t-shirts I can wear, I want to multiply each, not multiply them:

___4_______ x _____6______ = 24

# of shirts # of shorts

I do not want to add them, which would give me 10, the wrong answer, in this case.

The reason I bring up the fundamental counting principle is that some questions will actually combine combinations with the fundamental counting principle (though you’ll likely use the fundamental counting principle more often with permutation questions). To give you an idea of how a combinations can show up along with the fundamental counting principle, try the following question:

Mrs. Pearson has 4 boys and 5 girls in her class. She is to choose 2 boys and 2 girls to serve on her grading committee. If one girl and one boy leave before she can make a selection, then how many unique committees can result from the information above?

(A) 9

(B) 12

(C) 18

(D) 22

(E) 120

The first step in this problem is recognizing whether we are dealing with combinations or permutations. Since, I am ‘choosing’ from a larger group, in this case two separate groups, I want to use combinations. Remember: once we’ve chosen the 2 boys or 3 girls, the position within the committee doesn’t matter. That is, either you are in the committee or out of the committee (there are no gold medalists here!)

The next thing to notice with this problem is that of the original 9 students, 2 leave, one boy and one girl. So that leaves us with 3 boys and 4 girls. We want to choose two each. Therefore, we have to set up one combination for boys and one for girls.

For boys, we have 3C2 and for girls we have 4C2. This gives us: 3C2 = 3 and 4C2 = 6

The final step is once we’ve figured out the combinations above, we have to use the fundamental counting principle and multiply the total number of possibilities in a committee, not add them: 3 x 6 = 18, answer (C).

Many students get stuck on this step and wonder, why don’t I add them. It’s a good question, so what I want you to do is imagine that you have 3 shirts and 3 pants. How many different shirt-pants getups can you wear? Well, for each shirt there are 3 options of pants. Therefore, we multiply and get 9.

My advice is to try about 5 or 6 more combinations/permutations problems so that you can get the hang of it. With a little practice, you’ll able to deal with most of the problems the GRE hands you. Even if you miss a question—likely because it is very difficult—the fundamentals in this post should be enough to help you understand the explanation to that question, so that you can get a similar question right in the future.

*Editor’s Note: This post was originally published in May 2011 and has been updated for freshness, accuracy, and comprehensiveness.*

i m looking for the difference between permutation and combination. and this blog simply help me to understand the difference. thank u

“For boys, we have 3C2 and for girls we have 4C2. This gives us: 3C2 = 3 and 4C2 = 6”

Please can you explain this part? What does C mean? Where did the 2s come from?

Thank you

Hi Bert 🙂

Happy to help! These are examples of the notation “nCR” (“n choose r”) to describe a situation in which we are choosing r things from n total things. So, when we write “3C2” we are indicating that we are choosing 2 boys out of a total of 3 boys. Similarly, “4C2” means we are choosing 2 out of a total of 4 girls.

We can calculate nCr using the combinations formula:

nCr = n!/[r!(n-r)!]

When we plug in n = 3, r = 2 to calculate the combinations of boys, we see that 3C2 = 3. And when we plug in n = 4, r = 2 to calculation the combinations of girls, we find that 4C2 = 6.

Hope this helps! 🙂

Hi Chris, the answer is 18 – however u have marked the answer to be (A). Good insight to the problem. Thanks

Hi Sudeepta!

Thanks for pointing that out! We’ve fixed the typo so that other students don’t get confused 😀

It’s also great to hear that you like the post!

Happy studying!

The MAGOOSH word has two Os in it, so if we count the number of possible rearrangements, we use permutation divided by 2, yes?

Hi Phuong,

To count the total number of unique ways that we can rearrange the word “MAGOOSH,” we would divide the factorial of the total number of repeating letters (7!) by the factorial of the number of repeating letters (2!). So the answer would be 7!/2!. Since 2!=2*1=2, we would just divide by 2 in this case.

What if i want to make a password of 11 numbers from (0-9) 10 numbers with repetition allowed. i think the answer would be just 10^11. correct? i assume this is NOT a permutation nor a combination problem….

Hi Omar 🙂 You’re correct that the number of possible passwords would be 10^11, as each digit of the password could be any number 0 to 9. This example uses the Fundamental Counting Principle, where we multiply the number of possible options for each position together to find the total number of possible passwords. Since there are 10 possible options for each position, the number of possible passwords is 10^11.

very helpful! 🙂

nailed it.

wooohoooo!

What is “C” in your formulas. I believe you expect the reader to know – so you do not go over this step. please help

” If *two girls* leave before she can make a selection ” , “The next thing to notice with this problem is that of the original 9 students, 2 leave, one boy and one girl “.

Can’t resolve the ambiguity here .

Sorry for the confusion, Mat! That was a typo on my part :). I’ve changed the wording in the question so it is no longer ambiguous.

Thanks for spotting that!

Hey Chris!!

It is, 3C2 for boys and 4C2 for girls, we get 3 possible arrangements of boys and how is it 6 for girls? Because 4C2 should be 12 right? I’m confused!

Hi Ank,

Actually it is 6 possible arrangements for boys (4C2) and 3 possible arrangements for girls (3C2). This gives us 6 x 3 = 18.

Hope that clears things up :)!

Hi Chris,

Still not clear how boys 4C2 is 6 and not 12.could you please explain this part.

You got 4 boys and 2 of those boys will get chosen.

4*3=12. The 2 boys can be arranged in how many ways?

2*1=2. So 12/2=6. Hope that’s helpful

yup

4C2 stands for 4Combination2 formula === n!/(n-r)!.r!

Here n is 4 and r is 2.

Lets calculate: 4!/(4-2)!.2! = 4!/2!.2! = 24/4 = 6.

Hope u got it now

According to my calculations, the answer comes to 46…

There will be 3 scenarios

1. 2 boys leave

2. 2 girls leave

3. 1 girl and 1 boy leaves

In this case, the answer would be

(2C2 x 5C2) + (4C2 x 3C2) + (3C2 x 4C2)

Is this right?

Hi Nikhil,

You are definitely right. However, I already noted this ambiguity in the comments right above your original posting. Sorry if the problem was confusing.

Suppose the question appears as given here, i.e. without the sex of the leaving students being mentioned, then how would it affect the solution?

I think the question is a little bit nebulous. The sexes of the two students that left was not stated in the question, It could have been two girls or two boys. Then the answer would have been 18 or 10 respectively

Thanks for your sharp eyes – you are right indeed! The question is nebulous because the answer would change depending on whether two girls or two boy left (for that matter there is nothing explicitly stating that the two who left were of the same gender).

In the explanation, I mention that one boy and one girl leave. For some reason that information never made it into the original question. Again thanks for spotting the oversight!

The problem can be solved by deducting two females.

Boys:

4 total boys, so 4!

2 boys are chosen, 2 are not, therefore –

4!

–

2!x2!

=

4x3x2x1/2x12x1 = 6

Girls:

5 -2=3 total girls, so 3!

2 girls are chosen = 2!, and 1 was not = 1!

3!

–

2! x1!

=

3x2x1/2×1 = 3

6×3 = 18, answer is C.

Correction –

The line, “4x3x2x1/2x12x1 = 6” should read, “4x3x2x1/2x1x2x1 = 6”

Yes, that is the exact method! Also, I changed the wording in the problem to get rid of any ambiguity.