The number of ways of choosing a committee of two women and three men

Miss C is taken
(A) B included =>4C1​×4C2​=24
(B) B excluded⇒4C1​×5C3​=40
(ii) Miss C is not taken
⇒B does not come; ⇒4C2​×5C3​=60
⇒Total=124

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Solution:

There are 3 cases to consider: 1) Mr. A is on the committee, but not Mr. B, 2) Mr. B is on the committee, but not Mr. A, and 3) neither Mr. A nor Mr. B is on the committee. Let’s now analyze each case:

Case 1: Mr. A is on the committee, but not Mr. B.

If Mr. A is on the committee, then we only need to choose 2 more men from the 4 remaining men since Mr. B can’t be on the committee. Since we also need to choose 2 women from the 5 women, the number of ways this can be done is 4C2 x 5C2 = 6 x 10 = 60.

Case 2: Mr. B is on the committee, but not Mr. A. (and Miss C is on the committee)

If Mr. B is on the committee, then we only need to choose 2 more men from the 4 remaining men since Mr. A refuses to serve with Mr. B. Furthermore, since Miss C also needs to be on the committee, there is 1 more woman to choose from the 4 remaining women; the number of ways this can be done is 4C2 x 4C1 = 6 x 4 = 24.

Case 3: Neither Mr. A nor Mr. B is on the committee.

If neither Mr. A nor Mr. B is on the committee, then we need to choose 3 men from the remaining 4 men. Since we also need to choose 2 women from the 5 women, the number of ways this can be done is 4C3 x 5C2 = 4 x 10 = 40.

Therefore, the total number of ways this can be done is 60 + 24 + 40 = 124.

Answer: C