The title of this post may seem facetious. After all, even the person most allergic to math, most traumatized by math, still remembers how to count! The GMAT, of course, generally will not ask you, for example, to count from one to seven. The GMAT may give you a more complex scenario, and ask you to count how many ways can such-and-such happen. For example

1) Shakespeare wrote fifteen comedies (including the so-called “romances”), ten histories, and twelve tragedies. If a summer Shakespeare festival always has one comedy, one history, and two tragedies, how many different combinations of plays can the festival host?

As you see, this “counting” is a little more challenging than the kind of “counting” you learned in your salad days. I would like to convince you, though, that you are quite capable of solving problems like this.

## The Fundamental Counting Principle

This one big idea will give you a lot of mileage on any of the problems where the GMAT asks you to count things.

If option #1 has P alternatives and option #2 has Q alternatives (assuming that the two sets of alternatives have no overlap), then total number of different pairs we can form is P*Q. For example: Shakespeare wrote fifteen comedies and ten histories. If we want to select one comedy and one history, the total number of possible pairs is 15*10 = 150.

The FCP easily extends from two choices to three or any higher number. However many collections of alternatives there are, you simply multiple the number of alternatives in each set to produce the total number of combinations. For example: Shakespeare wrote fifteen comedies, ten histories, and twelve tragedies. If we are going to pick one of each kind, and ask how many different trios of plays can we create, the total number is simply 15*10*12. BTW, figuring that out without a calculator is not so hard. First, on the 15*12, use the “doubling & halving” trick —– 15 times 2 is 30, and 12 divided by 2 is 6, so 15*12 = 30*6 = 180, and last, the easiest of all, multiplying by ten: 15*12*10 = 1800.

## Permutations and Combinations

Before we can answer the original question posed, we have to clarify some terminology about counting. A **permutation** is a set in which order matters —- AABC is a different permutation from BACA. A **combination** is a set in which order does not matter —- AABC and BACA are the same combination of four letters.

This distinction is important in counting because we have to know whether to include the sets that repeat elements in different order. In question #1 above, the question explicitly asks for combinations. In other words, if we pick *Hamlet* and then *King Lear*, that’s will be considered the same as picking *King Lear* and then *Hamlet*.

This post: http://magoosh.com/gmat/2012/gmat-permutations-and-combinations/ considers permutations and combinations in greater detail. For the purposes of this post, we just have to be careful to consider what we are counting.

## How Many Pairs of Tragedies?

Shakespeare wrote twelve tragedies, and we want to pick a pair from these. How many different pairs can we pick? We will use the FCP.

Let’s break the task into a first choice and a second choice. For the first choice, we can choose any of the 12 tragedies, so we have 12 choices. Suppose, for the sake of argument, we pick *Romeo and Juliet* on the first choice. Now, for the second choice, notice that we don’t still have 12 choices, because we don’t want to pick *Romeo and Juliet* again on the second choice. In general, on the second choice, you have one fewer choice than you had on the first choice, because you don’t want to duplicate whatever element was chosen the first time. Thus, on the second choice there are eleven choices. By the FCP, it would seem the total number of possible pairs of would be 12*11.

If we were interested in permutations, then 12*11 would be the correct answer. Here, though, we are interested in combinations, not permutations. Again, if we pick *Romeo and Juliet*, then *Macbeth*, we will count that as the same pair as *Macbeth* first, followed by *Romeo and Juliet*. The figure of 12*11 automatically counts each permutations, and so counts every pair twice, as AB and then as BA. Thus, we have to divide 12*11 by two. Thus, 12*11/2 = 6*11 = 66 is the number of possible pairs of tragedies.

## The Whole Shebang

Now, we can answer the entire question. We want the number of combinations of four plays consisting of one comedy, one history, and two tragedies. By the FCP, that’s 15*10*66. Notice the math tricks to multiply easily without a calculator.

Split 66 back into 6*11 —- 15*10*6*11

Switch the order —– 15*6*11*10

Use doubling-halving on the 15 & 6 —- (15*6)*11*10 = (30*3)*11*10 = 90*11*10

Then 90*11 —- 90*11*10 = 990*10

Then, the easiest, multiplying by ten — 990*10 = 9900

Thus, the festival can come up with 9900 combinations consisting of one comedy, one history, and two tragedies.

## Practice Question

Here’s a practice question on this very topic: https://gmat.magoosh.com/questions/845

In the Shakespeare’s problem, isn’t it required to divide 15*10*66 by 3!, since we just want the number of sets of the four plays and not the order in which the four plays are hosted?

Hi, I am wondering the same. Was this question answered?

Excellent question! No, we do not need to divide the product 15*10*66 by 3! in this question. It’s definitely tempting to say that “order doesn’t matter”, because a comedy and a history is the same as a history and a comedy, but let’s try making those 3 tasks:

pick a comedy, pick a history, pick a tragedy

or

_____ , _____ , _____

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

comedy, _____ , _____

That’s NOT the same as

_____ , _____ , comedy

A comedy in the first blank is not the same as a comedy in the last blank, because the last blank is for choosing a tragedy and a comedy cannot be a tragedy. For that reason, we use the FCP to determine the total number of different combinations of plays that can the festival host.

I hope this helps

Hi Mike,

In regards to the example, does the combinations formula automatically consider repeated cases?

12C2 is equal to the thinking of having 12 tragedies for the 1st choice followed by 11 tragedies for the 2nd choice. But since a choice of A&B is the same as B&A the total number of tragedies must be divided by 2.

So the forumla 12C2 when completed automatically removes the repeat cases?

Dear Kelvin,

That’s a great question, and the answer is subtle. The combination formula does automatically take into account the

repetitions that come from different orders of selection. If we had a set of 12 distinct items, and were going to select two at random, the number of pairs would be 12C2 = 66. The fact that we could pick {A, B} as A then B, or as B then A, is already included in the 12C2 formula, and we need make no adjustment to the end result.Now, having said that, the formula is not immune to all kinds of repetitions. Suppose we had a pool of twelve that consisted of three identical items and 9 other distinct items—there we would get repetitions from the multiple copies of the repeated item. The basic combinations formula does NOT account for that kind of repetition, and the number of distinct pairs would not be 12C2. Let’s say the pool were 3 A’s, and then B-J. There are ten distinct items in the pool, which could give us 10C2 = 45 pairs of two distinct items; in addition, we could also pick the pair {A, A}, for a total of 46 pairs.

Does all this make sense?

Mike

Mike,

Thanks for a great article on a tricky subject.

Question:

How would you structure the solution if you wanted 3 tragedies (or any set of 3)?

Thanks,

Jake

Jake,

You’re welcome! I’m glad you found it helpful! If you wanted to know how many sets of 3 tragedies from the 12, that would be the combination 12C3. I would urge you to read this post on combinations:

http://magoosh.com/gmat/2012/gmat-permutations-and-combinations/

and this post on calculating combinations:

http://magoosh.com/gmat/2012/gmat-math-calculating-combinations/

Does all this make sense?

Mike

Permutations and Combinations both seem like such difficult concepts but when you lay it out like this and actually explain why we need to do what we do, it becomes so much simpler. Thanks for such an informative post. I’m loving this blog.

Dear Ryan,

Thank you for your kind words. Best of luck to you.

Mike