Dec
26
2012

# Never Fear, A Combinations Shortcut is Here!

Combinations is among the most feared of the math topics that can appear on the GRE quantitative section. I suppose that this makes sense – while many of us probably do some basic arithmetic in our everyday lives, it’s not often that we have to figure out how many different groups of 3 candles we could pick from a total of 7 different candles (I vote to eschew the “picking” and just light all of them, especially if they’re scented).

As you’ve been studying, you’ve likely stumbled up on the following formula for combinations: , where n is the total number of items in the set and k is the number of items being chosen. This formula will certainly get you the correct answer, but as with so many things GRE quant-related, there is a faster way.

Instead of using the full formula, you can use this simplified, but still effective, version:

# of combinations =  n  *   n-1 * n-2  …   x

k      k-1     k-2        1

Let’s put this into context in a real example:

This is an example of a multi-group combinations problem, and it’s the most common twist that the test-writers will add to a combinations problem to increase the difficulty level. To solve these problems, begin by finding the number of combinations for each group separately. We’ll start with the seniors: There are 8 total, and we want to pick 4 of them. So our n is 8, and k is 4. If we apply the approach discussed above, we’ll begin by putting n over k:

8

4

We’ll continue creating new fractions, that will all be multiplied together, by subtracting 1 from the numerator and 1 from the denominator in each new fraction until we hit a denominator of 1. In this case, that looks like:

8 * 7 * 6 * 5

4   3    2   1

Before you start multiplying, cancel out any common terms from the numerator and the denominator. Here, the 6 on top cancels out the 3 and 2 on bottom, and 8/4 simplifies to 2. So we’ve cancelled out all of the numbers on the denominator, and we’re left with 2*7*5 = 70: There are 70 possible combinations of seniors.

We still need to figure out how many groups of 2 we can make out of the 16 juniors, so let’s set up another set of fractions:

16 * 15

2       1

This reduces to 8*15, or 120.

Now that we’ve figured out the number of combinations of juniors and the number of combinations of seniors, we just need to multiply the two results together: 70*120 = 8,400. Answer choice (C) is correct.

With a solid understanding of how to find a number of possible combinations in the simplest possible way, and an example of how to tackle multi-group combinations questions, it’s almost as if the holidays have come early! Keep up the hard work that you’ve been putting in this year, and let us know if you have any questions in the comments.

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#### About the Author: Teresa Rupp

Teresa Rupp has been a Kaplan GRE teacher since the beginning of 2010. She graduated from Georgetown University with a degree in Middle Eastern Studies, which has left her with an enduring love of Lebanese cuisine. When she’s not coaching students to Test Day success in Baltimore and in Kaplan’s Anywhere classes, Teresa can usually be found reading, doing crossword puzzles, or hiking with Piper, her Welsh Springer Spaniel (who also enjoys Lebanese food).

• Dinesh Purohit

I like it, you have made this very simple. Thanks.

• Teresa Rupp

I’m glad that you liked it, Dinesh! Combinations has a reputation for being difficult, but it’s just as tackle-able as any other topic that you’ll see on the GRE if you have a consistent, strategic approach.

• Anil Pandey

Thank you. I like this trick, it makes it look so simple, however, one thing always intrigues me. How many permutation combination problem is a test taker supposed to take in maximum at a GRE test?

• Katie

But what do you do if everything in the denominator doesn’t cancel out?

• Alex

Awsome thanks so much for this really saved me Teresa

• Tony

In the part with 16 * 15 and 2 1…where did the 15 come from?

• Ingrid

Tony, you subtract one from the denominator and one from the numerator until the denominator is a 1.