When trying to decide whether to use the combination formula, many students have trouble applying the “does order matter” test–deciding if outcomes differ. It’s easy to say order doesn’t matter when it in fact does, and vice versa. Once you get it, it’s really useful, but it can take a bit of thinking and re-thinking.

Let’s look at a couple of examples to illustrate.

## A Simple Case of the FCP

Suppose we have 5 people waiting for 3 seats. Now let’s say I want to count how many ways 3 people can be arranged in those seats. There are three tasks here–one for each seat.

*fill 1st seat, fill 2nd seat, fill 3rd seat*

We can write those as 3 blanks:

_____ , _____ , _____

With this done, we’re most of the way there. We just need to check if filling one seat with one person is the same as filling another seat with that same person.

Amy , _____ , _____

_____ , Amy , _____

Now, we know those aren’t the same, because that makes two different line-ups. Keep in mind which task each blank represents.** ***Amy in seat 1 is not the same as Amy in seat 2*** .** So this is

*not*a combination. Instead, we can use the fundamental counting principle (FCP).

5 options , 4 options , 3 options

5*4*3 = 60.

If we were simply looking for a group of people–not paying attention to which person takes which chair, then we’d use a combination because the tasks aren’t inherently separate. There would be no significance of which blank represents which task. That might be asking “how many ways three people can be chosen for a committee from a group of five,” for example. In that case, we *can’t* draw the tasks as separate. Selecting the first person for the committee is the same as as selecting the second person. We could even choose all three at the same time, in one task–we don’t have to break it up into separate events. In that case, we have to use the combination formula as given in the our combination lesson videos.

5C3

5! / ([5-3]! *3!)

5! / ([2]! *3!)

5*4 / 2

10

## A Permutation that *Looks* like a Combination

Here’s another example. If you have 4 shirts, 4 pairs of pants, and 4 hats, and you choose one of each, is it a combination or does order matter? It’s tempting to say “order doesn’t matter”, because a t-shirt and jeans is the same as jeans and a t-shirt, but let’s try making those 3 tasks:

*pick a shirt, pick pants, pick a hat*

or

_____ , _____ , _____

Again, remember that each blank must represent one task alone, and we don’t move the blanks around. So if we say

red t-shirt, _____ , _____

That’s NOT the same as

_____ , _____ , red t-shirt

*A shirt in the first blank is not the same as a shirt in the last blank,* because the last blank is for choosing a *hat* and a shirt cannot be a hat. It’s not very fashionable at the moment, at least.

That means that we can use the fundamental counting principle here. First, we look at the shirt task. There are 4 shirts, so we have 4 possibilities. Next, and *separately, *we look at the pants. There are 4 pairs, so that’s another 4 possibilities for each shirt. So far, that’s 4*4. And finally, we have hats: there are another 4 hats, so that’s another 4 possibilities for each match up of shirt and pants. That’s 4*4*4.

So if we have *separate* blanks for each task, just fill in the blank with how many choices can be made in that specific step. Then multiply all the numbers for your answer.

On the other hand, if we were looking for ANY three pieces of clothing from the twelve total, we would use a combination formula. Again, this could be done in a single step–reach into a bag of clothes and grab three things. There’s no differentiation between them. This would give us 12C3, or 220.

Basically, this is all about drawing the blanks to represent the tasks. As you do so, ask if the item or person you’re picking can be moved around to different blanks without changing the situation. (Move the items–not the blanks!). If the item can be moved without changing the situation, it’s a combination–order doesn’t matter. If moving the item changes the situation (like in the line-up) or is impossible (like with the clothes), then we use the FCP.

P.S. Ready to improve your GRE score? Get started today.

Thanks. That clarifies few crucial things in a more visual way by using the example of blanks.

You’re welcome, Sourabh! I’m glad it was helpful.

This article confused me even more. Very confusing and examples aren’t well explained.

Hi Simran,

Combinations, Permutations and the FCP are some of the most difficult concepts on the GRE test. One of the best ways to improve your skills here is to keep reading these sorts of articles, puzzle through the questions and try to deepen your understanding. This might not happen overnight; it will likely take some time before you will feel completely comfortable with these types of questions. If you have any specific questions, please let us know and we will try to get back to you as soon as possible 🙂

Hello, by reading this article I can infer that FCP (Fundamental Counting Principle) is same as Permutation. Right?

Not quite, Rohit, although these two math principles (FCP and permutation) are closely related. A permutation is an ordering of

n things. So for example, say you want to know how many different ways you can order the digits 2, 4, and 6. 624 would be onepermutationof this order. 426 would be another permutation. You can also use “permutation” to refer to a complete set of all possible sequences of a group of things.The Fundamental Counting Principle, on the other hand, is a

methodused to determinehow manypermutations are possible in a set. In the case of permutations for 2,4, and 6, , we can use the fundamental counting principle to say the number of permutations n! = n*(n-1)*(n-2), since n = 3 (which is to say, three digits.). So, 3! = 3*2*1 = 6. So there are six possible individual permutations in the total 2/4/6 permutation set.In short, a “permutation” refers to a sequenced combination of things, a set of sequences. FCP refers to one method, often the best method, for determining the number of individual permutations in a set.

I still don’t understand how to figure out the answer to the question, “A Permutation that Looks like a Combination”. What is the answer? And how do you figure it out? I understand the idea on how it is not a combination, I just don’t understand how to solve the problem.

Thanks!

Hi there,

When we have a permutation, we use the following formula: nPr=(n!)/(n-r)!

So we are just applying this formula here! For each blank, we need to find 4P1, which becomes: 4P1=4!/(4-1)!

Which is 4P1=4!/3!

This becomes (4*3*2*1)/(3*2*1), the 3, 2 and 1 cancel from the numerator and denominator which leaves us with just 4.

In order to find all of the possible ways to combine an outfit, we need to multiple these together for the shirts, pants and hats. We are choosing one from four of each, so our permutation equation is always the same. The answer then becomes 4*4*4.

Does that make sense? You can learn some more about these specific methods here: https://magoosh.com/gre/2011/gre-math-combinations-and-permutations/

I solved this using combination as:

selection of the shirt can be done in 4C1 ways and similarly for the others thus making 4C1 * 4C1 * 4C1 which gives 4*4*4 ways

++correction

Can you please explain more on what you mean exactly by order. That’s the only part that confuses me in this wonderfully explained article.

Thanks.

Hi Ayesha,

What we mean by “order” is this question: If I have ABC is that considered different than, for example, CBA?

If I am deciding which three friends to invite to my party, sending invitations to ABC is the same as sending them to CBA. All three of my friends are going to come to the same event. In contrast, if I am deciding the order my friends will present their projects in class, then ABC is a different reality than CBA. In each of these, my friends have a different place in line (first, second, third) and thus the order matters.

I hope that clears it up! 🙂

might be vague question but when presenting the project, ultimately all three are presenting project because even though they are presenting in different positions, their presentation is independent of each other.

Please clarrify

I see what you’re getting at! To clarify, order

couldmatter in the case of three different people presenting a project, because it’s possible for those three people to present the project at different times/in different orders. But even in a case like that, order might not matter if the GRE story problem doesn’t actually concern order. Suppose, for instance, that a story problem indicated each presentation took a different length of time and asked you to solve for a variable, with a format that might look like this:Presentation A is 1.2 times longer than presentation B. Presentation B is 2 times shorter than presentation C, and the total combined time of all three presentations is 6.5 hours. How long is presentation C?

In this case, order wouldn’t matter, because the presentations really are being presented independently of each other where order is concerned. They could be happening simultaneously in different rooms, or on completely different days. The problem doesn’t say anything about presentation order.

On the other hand, if the focus of the problem was the possible different orders the presentations could go in within a single class (as mentioned above), order DOES matter. Does that make sense?

Hi Lucas,

if I understand your examples correctly, the first one implies that choosing for example “red shirt, blue hat, green pants” is not the same as choosing “green pants, blue hat, red shirt” (simply bc of te order), while the second example doesn’t distinguish between the shirt, pants, hat at all.

So how do you account for different types but the order of picking them doesn’t matter, i.e. you re required to choose 1 shirt, 1 pair of pants and 1 hat but “red shirt, blue hat, green pants” and “green pants, blue hat, red shirt” is the same?

I thought it was (4*4*4)/3 but that doesn’t yield an integer and, thus, doesn’t make any sense. So is my understanding of your example 1 wrong or my calculation in this paragraph?

Thanks a lot!

btw I am referring to the “A Permutation that Looks like a Combination” part only.

Hi,

It doesn’t appear that Lucas ever responded to your question. ??? Here’s a response to your problem (wrt Lucas’ way of explaining it.)

When you write out the 3 blank spaces for shirts, pants, hat, these spaces do NOT represent an actual “place-ordering”. Each space is actually a “category” of clothing. Therefore your question about (t-shirt, jeans, baseball-hat) VERSUS (jeans, baseball-hat, t-shirt) makes no sense. Make sense? : ) The order you write the items in the blank spaces isn’t an actual place-order, but rather a category-order.

This is why Lucas wrote “When a permutation looks like a combination”. This problem only looks like a combination b/c list those three items in any order and it’s still the same overall outfit, just like a combination. But it IS A PERMUTATION, actually, b/c the #choices you put in each blank space must be the number of choices for the respective category. It’s an ordering (permutation) of category, not of place in line, sort of speak.

Make sense?

It really helps me, Thanks for your nice explanation.

Nice comparison, was really helpful. Thank u

You’re welcome!

It really helped me.Thank you.