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7 M and 6 W 5 to be selected atleast 3M So combination are (3M*2W) or (4M*1W) or (5M) = (7c3 * 6c2) + (7c4 * 6c1) + (7c5) = 525 + 210 + 21 = 756
Answer

Given 7M and 6W and to form a committee of 5

so the combinations are (3M +2W) or (4M+1W) or (5M) since atleast 3 M needed

=> ( 7C3 x 6C2 ) + ( 7C4 x 6C1 ) + ( 7C5)

=> 525 + 210 + 21

=> 756

option c

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