If you’ve been paying attention to the exciting world of GMAT prep, you know that GMAC released two new practice tests fairly recently. I’d mentioned in a previous post that I was going to write about any conspicuous trends I noted, and one unmistakable pattern I’ve seen with my students is that probability questions seem to be cropping up with greater and greater frequency.

While these questions don’t seem fundamentally different from what we’ve seen in the past, there does seem to be a greater emphasis on probability questions for which a strong command of combinations and permutations will prove indispensable.

First, recall that the probability of x is the number ways x can occur/number of total possible outcomes (or p(x) = # desired/ # total). Another way to think about this equation is to see it as a ratio of two combinations or permutations. The number of ways x can occur is one combination (or permutation), and the total number of possible outcomes is another.

Keeping this in mind, let’s tackle this new official prompt:

*From a group of 8 volunteers, including Andrew and Karen, 4 people are to be selected at random to organize a charity event. What is the probability that Andrew will be among the 4 volunteers selected and Karen will not?*

*(A) 3/7*

*(B) 5/12*

*(C) 27/70*

*(D) 2/7*

*(E) 9/35*

Typically, I’ll start by calculating the total number of possible outcomes, as this calculation tends to be the more straightforward one. We’ve got 8 volunteers, and we want to know the number of total ways we can select 4 people from these 8 volunteers. Note, also, that the order does not matter – group of Tiffany, Mike, Louis, and Amy is the same as a group of Louis, Amy, Mike, and Tiffany. We’re not assigning titles or putting anyone in seats, so this is a combination.

If we use our combination formula N!/[(K!*(N-K)!] then N, our total pool of candidates, is 8, and K, the number we’re selecting, is 4. We get 8!/(4!*4!), which comes out to 70. At this point, we know that the denominator must be a factor of 70, so anything that doesn’t meet this criterion is out. In this case, this only allows us to eliminate B.

Now we want our desired outcomes, in which Andrew is selected and Karen is not. Imagine that you’re responsible for assembling this group of four from a total pool of eight people. You plan on putting your group of four in a conference room. Your supervisor tells you that Andrew must be in and Karen must not be, so you take Andrew and put him in the conference room. Now you’ve got three more spots to fill and seven people remaining. But remember that Karen *cannot* be part of this group. That means you only have 6 people to choose from to fill those other 3 spots in the room.

Put another way, think of the combination as the number of *choices* you have. Andrew and Karen are not choices – you’ve been ordered to include one of them and not the other. Of the 4 spots in the conference room, you only get to choose 3. And you’re only selecting from the other 6 people for those spots. Now N = 6 and K = 3. Plugging these into our trusty combination formula, we get 6!/(3!*3!), which comes out to 20.

Summarizing, we know that there are 20 ways to create our desired group of 4, and 70 total ways to select 4 people from a pool of 8, giving us a probability of 20/70, or 2/7, so the correct answer is D.

Takeaway: Probability questions can be viewed as ratios of combinations or permutations, so when you brush up on combinatorics, you’re also bolstering your probability fundamentals. Anytime you’re stuck on a complex probability question, break your calculation down into its component parts – find the total number of possible outcomes first, then find the total number of desired outcomes. Like virtually every hard question on the GMAT, probability questions are never as hard as they first seem.

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*By David Goldstein, a Veritas Prep GMAT instructor based in Boston. You can find more articles written by him here.*