Suppose that a department contains 13 men and 19 women. How many different commi

Question

Suppose that a department contains 13 men and 19 women. How many different committees of 6 members are possible if the committee must have strictly more women than men?

Explanation / Answer

Break it down into 3 cases: CASE 1: Committee with 6 women (no men) Choose the 6 women --> C(19,6) = (19 x 18 x 17 x 16 x 15 x 14) / (6 x 5 x 4 x 3 x 2 x 1) = 27,132 ways CASE 2: Committee with 5 women and 1 man Choose the 5 women --> C(19,5) = 11,628 ways Choose the 1 man --> C(9,1) = 9 ways Multiply: C(19,5) * C(9,1) = 104,652 ways CASE 3: Committee with 4 women and 2 men: Choose the 4 women --> C(19,4) = 3,876 ways Choose the 2 men --> C(9,2) = 36 ways Multiply: C(19,4) * C(9,2) = 139,536 ways There aren't any other valid cases (e.g. 3W+3M, 2W+4M, 1W+5M, 6M) so you are done. Add up all the valid cases: 27,132 + 104,652 + 139,536 = 271,320 ways Answer: There are 271,320 ways to form a committee of 6 members (where there are strictly more women than men).

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