Question
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?
Answer: Option D
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We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).
So, Required number of ways = = (7C3 x 6C2) + (7C4 x 6C1) + (7C5)
= \(\left(\frac{7\times6\times5}{3\times2\times1}\times\frac{6\times5}{2\times1}\right)\) + (7C3 x 6C1) + (7C2)
= \(525+\left(\frac{7\times6\times5}{3\times2\times1}\times6\right)+\left(\frac{7\times6}{2\times1}\right)\)
= (525 + 210 + 21)
= 756.
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