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## Backwards GMAT Combinations

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Most experienced GMAT students have encountered "combinations" problems, and have at least an idea of how to solve them. For the unfamiliar, these are questions of the following sort:

A three-person committee is to be selected from among the employees of a six-person firm. If any employee of the firm can be selected to be on the committee, how many distinct committees can be formed?

While there are twists that the GMAT can apply to these questions, they usually follow this format: a certain number of people (or items) that can be part of the group, and a certain size of the desired group. However, there are a handful of questions that leave the size of the group (6, in the example above) unknown, and provide the number of combinations.

Combinations Backwards

I included a question like that in last week's set of five free newsletter problems. Here it is:

Two members of a certain club are selected to speak at the next club meeting. If there are 36 different possible selections of the 2 club members, how many members does the club have?

In this case, we have the subset of 2 members, and we have the total number of combinations.

On other types of questions, this would be a trivial change. In a rate question, it doesn't matter whether we are given rate and time, distance and time, or rate and distance: Given any pair of values, we can find the third. Since the combinations formula includes factorials, though, that makes things more difficult.

To solve, start with the combinations formula:

C = (n!) / (k!(n-k)!)

We know that C = 36, k = 2, and we're solving n. That gives us:

36 = (n!) / (2!(n-2)!)

Without some college math under your belt, you've probably never encountered anything like this before.

Simplifying Factorials

The factorial notation, n!, represents something more specific:

n! = n(n-1)(n-2)...(1)

We just don't know the value of n, so we don't know how many terms there are. However, we can simplify it a number of intermediate ways:

• n! = n(n-1)! (n times the factorial of n-1)
• n! = n(n-1)(n-2)! (n, times n-1, times the factorial of n-2)
• n! = n(n-1)(n-2)(n-3)! ...etc.

In this case, we know what we're trying to simplify, with a (n-2)! in the denominator. So substitute the second of those options for n!:

36 = (n(n-1)(n-2)!) / (2(n-2)!)

Now cancel out the (n-2)!:

36 = n(n-1) / 2

I'll leave the rest of the solution to you. You're not quite home free, but the combinations part of the problem is dealt with.

Generalizing the Solution

This type of math can get very complicated, and that's a signal that the GMAT (which knows you probably don't have a math degree) won't take it much farther. In fact, I've never seen a question like this that had a value of k other than 2.

Any time k=2, the combinations formula simplifies to the following:

C = n(n-1) / 2

So, if there are 8 tennis players in a league and we want to know how many possible pairings there are, we multiply 8 times 7 and divide by 2. (Try it with the original combinations formula: you'll see that it simplies to exactly that.)

Intuitively, that makes sense. In the tennis example, every one of the eight players has seven possible opponents. That makes 56 (8 times 7) possible pairings, but each pairing has been counted twice. (Joe playing Keith is the same as Keith playing Joe.) So we divide by 2.

If you know that, you don't have to do nearly as much work to get to the answer of the example we started with. Knowing that k=2, you could plug in the value for C:

36 = n(n-1) / 2

It's worth remembering this. A disproportionate number of "typical" combinations problems have k=2, and the "backwards" combination problem will almost always have k=2. If you understand it intuitively, that's even better.

About the author: Jeff Sackmann has written many GMAT preparation books, including the popular Total GMAT Math, Total GMAT Verbal, and GMAT 111. He has also created explanations for problems in The Official Guide, as well as 1,800 practice GMAT math questions.

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