How to Actually Do Combinatorics Problems on the GRE
Combinatorics—it’s a word none of us can say and none of us had ever heard of before we started studying for the GRE. It’s a fancy word that just means “the number of possibilities” or “all the ways something could go” (my definitions).
And once we do start to tackle it, we’re often stumped. Most books and online study tools teach us a formula that is deceptively simple. “Great!” we think. “I’m set!”
Then we get to a combinatorics problem on a practice test, and what happens? We get it wrong. The formula doesn’t work.
It’s a frustrating experience, and one I’m going to solve for you in this post.
First, though, a caveat—there are very few of these problems on the test. It’s not worth your time to spend hours studying combinatorics over, say, algebra, or percentage word problems. So while I want you to be empowered to answer combinatorics questions, I also don’t want you to overinvest your time at the expense of working on other areas that you’re likely to be tested on…like geometry.
GRE Combinatorics Strategies
Got it? Okay. Now let’s cover the combinatorics strategy that we teach at Manhattan Prep because it’s actually usable on GRE combinatorics problems. It’s not a simple formula, but unlike the simple formula, it’s applicable on the test.
First, two key terms: slots and labels.
Slots – a blank line for every decision
Labels – description of the category of item that will fill the slot
Here’s your process:
Step 1: Create slots for every decision you’re making.
Step 2: Label those slots according to the type of thing that will fill them.
Step 3: Enter the number of options for each slot.
Step 4: Multiply to find the total number of possibilities.
That’s (almost) it! We are 90% of the way done.
Let’s take a look at an example.
Suppose we are creating a singing group of four people, choosing from a group of six singers. Each person will have an assigned role: soprano, alto, tenor, and bass. We make four slots for each of the parts, and label them:
Now, we fill the slots with the number of options for each decision. For the first spot, we are choosing from 6 people:
For the second, we’re now choosing from 5 people—one person has already been chosen:
Let’s finish filling them in:
From here, we are almost done but not quite. We must consult our labels and ask: Are all of the labels different, or do any repeat? If they’re all different—which they are in this case—we’re done! We simply multiply to find the total number of possible book groups:
6 x 5 x 4 x 3 = 360 possible combinations
But what if we have repeated labels? In this case, we add one more step:
Step 1: Create slots for every decision you’re making.
Step 2: Label those slots according to the type of thing that will fill them.
Step 3: Enter the number of options for each slot.
Step 4: Multiply to find the total number of possibilities.
Step 5: Ask—do any of the labels repeat? If no, you’re done. If yes…
Say, for example, that we’re just making a singing group with no assigned roles? So the labels aren’t Soprano, Alto, Tenor, and Bass, they’re just Singer (I’ll use “S”):
Now, we have the same label—S—used more than once. Whenever you have a label used more than once, you must take one last step: divide by the factorial of the number of repeated labels. (The reason for this is that we have overcounted—360 is going to be too big. If you want a challenge, see if you can figure out why!)
In this case, we have four repeated Ss, so we will divide by 4 factorial:
And that’s our method!
GRE Combinatorics Strategy Review
Let’s review:
Step 1: Create slots for every decision you’re making.
Step 2: Label those slots according to the type of thing that will fill them.
Step 3: Enter the number of options for each slot.
Step 4: Multiply to find the total number of possibilities.
Step 5: Ask—do any of the labels repeat? If no, you’re done. If yes, divide by the factorial of the number of repeated labels.
One last thing—if you have multiple repeated labels, let’s say 4 S’s and and 3 Y’s (just randomly picking letters here), just divide by the factorial of both: 4! x 3!, or: 4 x 3 x 2 x 2 x 3 x 2 x 1.
Happy combinatorics-ing!
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Mary Richter is a Manhattan Prep instructor based in Nashville, Tennessee. Mary is one of those weirdos who loves taking standardized tests, and she has been teaching them for 15 years. When she’s not teaching the LSAT or GRE for ManhattanPrep, she’s writing novels under the last name Adkins. You can find them wherever you buy books. Check out Mary’s upcoming GRE prep offerings here!