Help! I Can’t Handle GMAT Probability and Combinatorics (Part 2)
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In the previous article in this series, we introduced two big ideas about GMAT probability and combinatorics:
- Most people find them counterintuitive.
- The best way to get past that is to list the possibilities.
In this article, we’ll focus more on #2. How do you list out the possibilities in a GMAT probability or combinatorics problem? Let’s try it on a simple probability problem.
If you flip three fair coins, what is the probability of getting exactly two heads?
Student A reasons as follows: “There are four different scenarios. You could get no heads, one head, two heads, or three heads. Therefore, the probability of getting two heads is one out of four, or ¼.”
Unfortunately, Student A is incorrect. That’s because Student A’s four scenarios aren’t equally likely. It’s actually more likely that you’ll get two heads than three heads. To see why (and to solve this problem), break down the possibilities more finely. Actually write out which coins will land heads up, and which ones will land tails up.
In three out of the eight cases, there are exactly two heads. The probability is 3/8.
How can you be sure that you’ve considered every case? The trick is to stay organized. To create the table above, I thought like this:
- Let’s start with three heads. There’s only one way for that to happen. List it out.
- Now, two heads. In that case, there’s one tail. The one tail could be the first coin, the second coin, or the third coin. List out those cases.
- Next, one head. That single head could be the first coin, the second coin, or the third coin. That’s three more cases. List them out.
- Finally, there could be zero heads. That’s the last case.
I’m sure that I listed every case, because I considered all of the possibilities, from ‘all heads’ down to ‘no heads.’ But actually, that’s not the only way to list your cases! Here’s a second way to get the same result.
This time, start by listing out the cases that start with two heads.
Then, consider the cases that start with two tails:
Then, consider the cases that start with a head and a tail:
Finally, the cases that start with a tail and a head:
There are still eight cases, and three of them still have exactly two heads. Once again, we can conclude that the probability is 3/8.
Before you keep reading, try listing out all of the scenarios that fit the following description. Stay organized, and use visuals if they help you!
A family of five is going on a road trip. Their car has five seats: a driver’s seat and one passenger seat in the front, and two window seats and a middle seat in the back. The family consists of Mom, Dad, and three kids: Alice, Bob, and Claire. Only Mom and Dad are able to drive. Mom can’t sit in the back, because she gets carsick, and Alice and Claire can’t sit next to each other, or else they’ll fight. In how many different ways can the family sit in the car?
Got your list? Okay, I’ll share how I thought about this one.
Let’s start with the driver, since that position has the fewest possibilities. Either Mom or Dad is driving. So, the first thing I jotted down on my paper was a two-column chart:
If Dad is driving, then Mom must be in the front passenger seat. She can’t be in the back, and that’s the only seat left. That leaves the three kids in the back.
If we put the three kids in the back, how many ways can we arrange them? Well, Bob has to be in the middle – otherwise, Alice and Claire would be next to each other, and we can’t have that. So there are really only two ways to do it. Either Alice is on the left and Claire is on the right, or it’s the other way around.
There are two possibilities, and they look like this:
What have we just learned? First, if Dad is driving, there are only two different ways to seat everyone else. Second, despite my GMAT score, I’m not so great at drawing cars.
Okay, let’s turn our attention to the second scenario. What if Mom is driving?
Anybody could be in the front seat, so let’s split it up again to make things simpler. Let’s suppose that Dad is in the front seat. How many ways could we arrange everybody else?
If Dad is in the front with Mom, we have the same two cases we already looked at. Alice and Claire can’t be next to each other, so Bob is in the middle and his sisters are on either side.
What if Dad isn’t in the front? Let’s try putting poor Bob in the front, instead of sticking him in the middle seat.
Dad would have to be in the middle, to separate Alice and Claire. That gives us two more possibilities.
What if Alice is in the front? Well, it doesn’t matter how we arrange the back seat, since there’s no risk of Alice and Claire fighting. There are 6 ways to arrange it, and they all work fine. The same thing happens if we put Claire in front: since one of the sisters is in the front, it doesn’t matter what order the back seat is in.
If Mom is driving, there are 2 + 2 + 6 + 6 = 16 different ways for the family to sit. Add the two cases where Dad is driving, and that makes a total of 18 possibilities.
Here’s what I really want you to notice. This process isn’t magic, and it isn’t math. It’s just about organization.
- First, we split up the problem to make it easier. This is one of the basic principles of counting cases.
- We split it up that way for a good reason. There were only two different people who could drive, so splitting up the problem like that made it easy to stay organized. If we’d split up the problem based on who was in the passenger seat (for example), we still would’ve gotten the right answer, but it would have been much messier.
- When we realized that Mom driving still gave us a lot of different cases, we split up the problem a second time.
- We stayed organized by drawing a chart on our paper and filling it in.
This problem only had a small number of scenarios. There’s one big question remaining: can we use the same technique to handle much larger problems? We can. In two weeks, check out the final article in this series to learn how. And while you’re waiting, try this practice problem:
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.) ?
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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.