Bunuel wrote:

Jane and Thomas are among the 8 people from which a committee of 4 people is to be selected. How many different possible committees of 4 people can be selected from these 8 people if at least one of either Jane or Thomas is to be selected?

A) 28

B) 46

C) 55

D) 63

E) 70

Kudos for a correct solution.

KAPLAN OFFICIAL SOLUTION:The first step in this problem is to recognize that order does not matter. We must create a committee of four people, but we are not putting these people in any sort of order. Because order does not matter, we must use the combinations formula, which is n!/[k!(n-k)!], in order to determine the number of possible outcomes.

However, as is the case with most advanced combinations problems, the solution is not as simple as plugging a few numbers into the formula. Instead, we need to consider exactly what types of outcomes we are looking for.

In this case, we need to select Jane and 3 other people, none of whom are Thomas; Thomas and 3 other people, none of whom are Jane; or Jane, Thomas and 2 other people. Additionally, we need to note that this is an ‘or’ situation, which means that we need to add all of these possibilities together.

First, if we select Jane and three others, none of whom are Thomas, we must pick Jane plus 3 out of the remaining 6 people. Because we can pick any 3 out of the 6, we set n = 6 and k = 3. When we plug this into the formula it gives us 6!/(3!3!) = 20 (Quick review: 3! = 3 x 2 x 1, 6! = 6 x 5 x 4 x 3 x 2 x 1.)

Next we determine how many ways we can select three people, none of whom are Jane, to go along with Thomas. This is identical mathematically, thus we end up with 20 in this case too.

Lastly, we determine how many outcomes are possible in which we have Jane, Thomas and 2 others. We now select 2 out of 6, so n = 6 and k = 2. 6!/(4!2!) = 15.

All we need to do to get the answer is add up all of the possible outcomes. 20 + 20 + 15 = 55, which is

the correct answer to this question (C).

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