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# Permutations and Combinations – Counting on the GMAT Doesn’t Have To Be So Difficult

All kinds of freaky formulas and rules to remember can make learning how to solve GMAT counting and combinatorics problems seem to be a major hassle. The truth is that by learning a little about how this all works, one can get this down no problem. So here is an outline that gets you to the more complex stuff while keeping it simple.

The entire thing is based on the fundamental counting principle, or FCP. The FCP states that when combining sets of choices we multiply the choices together to find the total number of possible combinations.

For instance, in a basic situation, if we have 2 choices of pants and 3 choices of tops, we can figure out how many possible outfits can be created by multiplying the number of pants choices by the number of tops choices.

Pants – 2 Choices

Tops – 3 Choices

2 x 3 = 6 Possible Outfit Combinations

We can confirm this and see how it works by writing out the combinations, using P1 and P2 as the pants choices and T1, T2 and T3 as the tops choices.

P1T1     P1T2     P1T3

P2T1     P2T2     P2T3

It’s pretty simple. For each of the 2 choices of pants there are 3 choices of tops.

This basic way of using the FCP can be applied to various situations with varying numbers of choices.

Here’s another situation.

On the menu of a restaurant there are 6 different appetizers, 10 different entrees and 5 different desserts. We can use the FCP to determine how many possible meals can be created.

For each of the 6 appetizer choices there are 10 entree choices, and for each of the 10 entree choices there are 5 dessert choices.

So the number of total possible combinations of meals is 6 x 10 x 5 = 300.

If you get that, you are well on your way to handling counting problems on the GMAT.

Ok now let’s try something else.

Let’s press a license plate. The plate has four slots and each of the outer two slots can be filled with any of the letters A through Z, and each of the inner two slots can be filled with any of the numeric digits, 0 through 9. So we have two outer slots with 26 choices for filling each of them and two inner slots with 10 choices for filling each of them.

Here’s an example of a plate that could be created. A17B.

There are 26 choices of letters to put in the first slot. For each of those there are 10 choices of digits to put in the second slot. For each of those there are 10 choices for the third slot. For each of those there are 26 choices for the fourth slot.

So the number of different plates we can create is 26 x 10 x 10 x 26 = 676 x 100 = 67,600.

That’s a lot of different plates, but the math is pretty straightforward.

Now let’s take this to another level.

We can have 10 books and a space on the shelf that can hold 3 of them. The question is how many arrangements of books can we put in those three spots. So now instead of repeating the choices as we could letters and digits in the license plate example, we have a bunch of choices, books that we can only use one time each. Once we put a book in the first slot, we can’t put that same book in the second or third slots.

Also, this is an arrangement problem where the order of the books matters. In other words if we change the order, we create a new arrangement, or permutation. The word permutation comes with images of formulas and concerns about complexity, but solving this is pretty simple. We just see how many choices of books we have for each slot and multiply the choices, using the FCP to see how many arrangements or permutations are possible.

We have 3 slots and ten books and we could fill the slots in any order, but let’s be simple about this. Let’s fill the slots from left to right.

To fill the first, the left, slot, we could choose any of the 10 books.

Now once a book has been chosen to fill that first slot there are only 9 left. So for each of the ten choices for  the first slot, we have 9 choices for the second slot.

After the second slot has been filled, for each of those 9 choices we are left with 8 books from which to choose to fill the third slot.

So to see how many arrangements we can come up with we just multiply the number of choices we had for the 3 slots.

10 x 9 x 8 = 720 different arrangements we can make by choosing 3 books from 10 and placing them in the three slots.

That is the basis of all GMAT permutation problems. I told you it doesn’t have to be complicated.

Now check this out. What if there were just three books to be arranged in the three slots? We need to understand this in order to see how to calculate combinations, where the order does not matter.

If we had just three books, there would be 3 to choose from for the first slot. For each of those three choices, there would be 2 left to choose from for the second slot and then just one left to pop into the third slot.

So three books in three slots can be arranged in 3 x 2 x 1 = 6 ways.

What if the order did not matter and we were to have 3 books to put into a box that holds 3 books? We would just put the three books into the box and be done. If order does not matter, then as far as we are concerned there is just one way to put three books into a box that holds three books.

Similarly if we were to have a box that holds 10 books and we didn’t care about the order, we would just put the books in the box. 10 books, 1 box, 1 way.

Now let’s look at the ten books we were choosing 3 from. Originally we cared about the order. So we were creating permutations of books. Now let’s just choose combinations of books. We just choose 3 books and put them into a box without being concerned about order.

When I was learning this I actually tried to figure out a formula myself and it was tough. So I looked it up and found out that it can be done pretty easily. Here’s what you do.

First you calculate the number of permutations of three books chosen from 10, as we did when putting books on the shelf, to come up with 10 x 9 x 8 = 720 arrangements.

Now check this out. Those permutations include arrangements of the various combinations of books. For every set of 3 books, there are multiple arrangements. All those arrangements add up to all the permutations. The thing is now we don’t care about all the arrangements of sets of 3 books. Each set of 3 is just considered 1 combination of books. So how do we get from the number of permutations to the number of combinations? We just divide.

From what we did earlier, we know that each set of 3 books can be arranged 6 different ways. Our set of permutations includes all of those ways. So to get from the set of permutations, where order matters, to the set of combinations, where order does not matter, we just divide by the number of arrangements of each combination of three books.

720 ÷ 6 = 120

So from 10 books, combinations of 3 books can be chosen 120 ways.

So that’s pretty much it. Sure, on the GMAT there will be more complex variations and other factors to deal with when solving counting problems, but once you get these basic principles, you are well on your way to rocking counting on the GMAT.