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.