Never Fear, A Combinations Shortcut is Here!
December 26, 2012
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:
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
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.