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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?

564
645
735
756
Explanation:

We may have [3 men and 2 women] or [4 men and 1 woman] or [5 men only].

$\therefore$ Required number of ways = (7C3 x 6C2) + (7C4 x 6C1) + (7C5)
=$ \left(\dfrac{7 \times 6 \times 5}{3 \times 2 \times 1} \times\dfrac{6 \times 5}{2 \times 1} \right) $+ (7C3 x 6C1) + (7C2)
= 525 +$ \left(\dfrac{7 \times 6 \times 5}{3 \times 2 \times 1} \times 6\right) $+$ \left(\dfrac{7 \times 6}{2 \times 1} \right) $
= (525 + 210 + 21)
= 756.
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