From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?

Answers (1)

At least 3 men are there on the committee.

Case - 1  3 men and 2 women = ^7C_3 \times ^6C_2 = \frac{7 \times 6 \times 5 \times 6 \times 5}{3 \times 2 \times 2} =525

Case -2  4 men and 1 woman = ^7C_4 \times ^6C_1 = \frac{7 \times 6 \times 5 \times 6 }{3 \times 2} =210

Case -3  5 men and 0 woman = ^7C_5 \times ^6C_0 = \frac{7 \times 6 }{2 } \times 1 =21

Total number of ways = 525 + 210 + 21 = 756

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