Help! I Can’t Handle GMAT Probability and Combinatorics (Part 3)
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In the previous articles in this series, we developed a critical skill for GMAT probability and combinatorics problems: listing out cases. Let’s start by taking another look at the practice problem from the end of the last article.
Six coworkers (Anil, Boris, Charlie, Dana, Emmaline, and Frank) are having dinner at a restaurant. They’ll sit in chairs that are evenly spaced around a circular table. Boris, Charlie, and Dana refuse to sit directly across from Anil, because he chews with his mouth open. Frank and Emmaline won’t sit next to each other. Finally, Emmaline and Dana insist on sitting next to each other. How many different arrangements will work? (Ignore the arrangements that come from ‘rotating’ the whole table – only focus on the relative positions of the diners.)
You know by now that you should solve problems like this by finding an organized way to list the possibilities. The key is to divide the problem into smaller, simpler problems, to make it easier to write that list.
In this problem, there are only two people who can sit across from Anil. Only Emmaline and Frank will put up with him! We’ll start by looking at cases where Emmaline sits across from Anil, and then we’ll look at cases where Frank sits across from Anil. Here’s what my scratch work looked like as I started this problem.
Next, I focused in on the table on the left. I know that Frank won’t sit next to Emmaline. So, Frank is either to the left of Anil, or to the right of Anil.
Also, Dana has to sit next to Emmaline. Combining those two facts together, there are four possible ways to set everything up. Here they are:
Finally, Boris and Charlie can sit wherever they’d like. That doubles the number of possibilities, giving us a total of 8.
Next, look at the scenarios where Frank sits across from Anil. Try it on your own: can you identify the 4 possible scenarios that work? Combining those with the 8 scenarios we found already gives a total of 12 ways to arrange the diners.
Now, that practice problem is more time-consuming than anything you’re likely to see on the GMAT. However, you can use the basic ideas from that problem on a wide range of GMAT probability and combinatorics problems!
On the GMAT, you may have heard about “order matters” and “order doesn’t matter” combinatorics problems—let’s talk a bit about those. Personally, I find the “order matters” terminology to be confusing, so I’m not going to use those words in this article. Instead, here’s a method that will use the skill of “splitting up the problem” that you’ve already been developing. Here’s an example problem.
In how many ways can a committee of 5 members be chosen from a class of 10 people?
(A) 40
(B) 126
(C) 252
(D) 6048
(E) 30240
Yikes—those numbers are too big to just list out the possibilities, like we have been. How do we start? Just like we have been, start by making the problem simpler. Don’t try to do the whole thing at once. Instead of asking yourself how to choose the committee, ask yourself: in how many ways can I choose the first member of the committee? Well, there are 10 people in the class, so there are 10 ways to do it.
Now, zoom in. If you’ve already chosen the first member, how many ways are there to choose the second member? There are 9 people left, so there are 9 ways to choose. For each of 10 first members, there are 9 second members. Keep doing this until you’ve chosen the entire committee: (10)(9)(8)(7)(6).
One of my high school math teachers loved to tell this joke:
How many legs does a cow have?
Eight: two front legs, two back legs, two left legs, and two right legs.
When we found the solution above, we made the exact same mistake. We counted each committee more than once, just like my math teacher counted each leg more than once. Let’s see how it happened.
We started by picking the first committee member. Let’s say that was Anil. Then, we picked a second committee member: that was Boris. Then, we picked Charlie, then Dana, then Emmaline. However, we separately counted the scenario where we started by picking Dana, then picked Anil, then Charlie, then Emmaline, then Boris. And we also separately counted the case where we picked Anil, then Charlie, then Boris, then Dana, then Emmaline. Those committees should actually all be the same committee, because they have the same people on them! We should have only counted it once. But we messed up—we counted it a bunch of different times, just like we counted each of the cow’s legs too many times.
However, you can fix this problem easily. You just have to divide by the number of times you counted each committee. (Notice that in my math teacher’s joke, we counted each leg twice—so, to get the right answer, you’d just have to divide by 2). How many times did we count the committee consisting of Anil, Boris, Charlie, Dana, and Emmaline? We counted it once for each possible order we could have picked them in. There are (5)(4)(3)(2)(1) different orders to put them in, which means we counted each committee 120 times. Our answer is 120 times too big.
Just divide by 120, and you have the right answer! Here’s what your scratch work would look like:
Okay, try it again, but more quickly:
How many different poker hands of 5 cards can be drawn from a 52-card deck?
There are 52 ways to draw the first card, 51 ways to draw the second card, and so on. That makes (52)(51)(50)(49)(48) different hands. Did we overcount? Yes! We counted each hand (5)(4)(3)(2)(1) = 120 times, just like we counted each committee 120 times. Divide by 120 to get your answer.
How about this one?
A teacher asks 3 students from a class of 7 to stand in a straight line from left to right. In how many different ways can this be done?
Okay, there are 7 ways to choose the student on the left, then 6 students who could be in the middle, then 5 students on the right. (7)(6)(5) possibilities. Did we overcount? Actually, we didn’t. We don’t have to divide. That’s because the line consisting of Anil, Boris, and Charlie should be counted separately from the line consisting of Charlie, Boris, and Anil. Those are two different lines, so you need to count them both individually. The right answer is (7)(6)(5) = 210—no division necessary.
Now you have a basic strategy for combinatorics problems with bigger numbers: count the possibilities as simply as possible. Then, divide if necessary to fix the “cow has eight legs” mistake. This strategy achieves the same thing as thinking about “order matters,” but it keeps you from having to worry about which type of problem you’re dealing with! Instead, you can just use math and common sense to figure it out. ?
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Chelsey Cooley
is a Manhattan Prep instructor based in Seattle, Washington. Chelsey always followed her heart when it came to her education. Luckily, her heart led her straight to the perfect background for GMAT and GRE teaching: she has undergraduate degrees in mathematics and history, a master’s degree in linguistics, a 790 on the GMAT, and a perfect 170/170 on the GRE. Check out Chelsey’s upcoming GRE prep offerings here.