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Few problems strike fear in the heart of test-takers quite like combinations problems. The usual arsenal of test-taking strategies–plugging-in, ball-parking, process of elimination–totally fails us when we tackle these questions.

The worst ones often include complications that make a difficult problem even more dreadful. Take this question for instance:

How many 5 person committees chosen at random from a group consisting of 3 men, 5 women, and 2 children contain at least 1 woman?

A) 250
B) 251
C) 252
D) 275
E) 300

What makes this problem difficult is the extra restriction that the committee must contain at least 1 woman. What we need is a powerful test-taking strategy that can help break up this complicated problem into smaller, simpler questions. Ideally, these “component parts” could then be joined together to allow us to indirectly calculate the original scenario. There is, in fact, a technique which can do exactly just that.

The key to solving this problem is realizing that the “Total” number of committees (committees with any number of women) is equal to the number of “Good” committees (those with at least 1 woman) plus the number of “Forbidden” committees (those with no women at all). In equation form, this would be:

C(Any Number of Women) = C(At Least 1 Woman) + C(No Women)

Upon re-arranging the equation, we see that:

C(At Least 1 Woman) = C(Any Number of Women) – C(No Women)

In other words, to find the number of committees with at least 1 woman, we can simply find the number of committees with any number of women, C(Any Number of Women), and subtract out the number of committees with no women at all, C(No Women). As it turns out, calculating the two components on the right side of the equation is far simpler than computing the value directly.

More generally, for combinations problems with complicated restrictions, we can often write the following equation:

C(Total) = C(Good) + C(Forbidden)

“Good” represents the desired combinations, “Total” represents all possible combinations, and “Forbidden” represents all combinations that are expressly forbidden. We can also re-arrange this equation to get:

C(Good) = C(Total) – C(Forbidden)

Now let’s return to our original committee problem and calculate the two sub-components:

To calculate the “Total” combinations, C(Total), we need to determine how many ways we can choose a 5 person committee out of a total of 3+5+2=10 people, ignoring gender and age. There are a total of 10 people, and we need to choose 5 out of that 10. Since all 5 members come from a single source (the group of 10 people), there is no replacement, and ordering does not matter, we can use the combinations formula: .

Calculating the “Forbidden” combinations is even simpler. If no women were allowed to be in the committee, we would need to find a 5 person committee out of the remaining 3 men and 2 children. In other words, we’d need to form a 5 person committee out of the remaining 3+2=5 people. With women excluded, the only 5 people left will all be in the committee, so there is only 1 way to form such a committee. (This informal observation agrees with the combinations formula, which would give C(5,5) = 1)

Using our equation, the number of committees with at least 1 woman is simply:

C(At Least 1 Woman) = C(10,5) – 1 = 251 (answer choice B)

When you take the GMAT, pay attention to the following hints that clue you in to using the Good = Total – Forbidden equation:

1. The question stem contains the key words “at least,” which adds a restriction that makes direct calculation burdensome
2. The “opposite” of the restriction (the forbidden scenario) is trivial to calculate

Many probability problems are simply minor variations of this same basic technique. The probability of the “Total” is simply 1, so that the equation reduces to P(Good) = 1 – P(Forbidden).

Now that you grasp this technique, you should have no trouble with mastering “at least” probability and combinations questions on the GMAT.

To review what you’ve learned, take a look at this week’s challenge problem:

A game at the state fair has a circular target with a radius of 10 cm on a square board measuring 30 cm on a side. Players win prizes if they throw two darts and hit only the circular area on at least one of the two attempts. What is the probability that Jim won the game?

A)
B)
C)
D)
E)

Find the solution and show your work in the comments below. In a few days, I’ll write up the answer for this problem.