Answer with Explanation:
We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).
Required number of ways = (7C3 x 6C2) + (7C4 x 6C1) + (7C5)
= ( 7 x 6 x 5 /3 x 2 x 1) x( 6 x 5 / 2 x 1 ) + (7C3 x 6C1) + (7C2)
= 525 +( 7 x 6 x 5 /3 x 2 x 1 x 6 ) + ( 7 x 6 / 2 x 1 )
= (525 + 210 + 21)
= 756.
We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).
Required number of ways = (7C3 x 6C2) + (7C4 x 6C1) + (7C5)
= ( 7 x 6 x 5 /3 x 2 x 1) x( 6 x 5 / 2 x 1 ) + (7C3 x 6C1) + (7C2)
= 525 +( 7 x 6 x 5 /3 x 2 x 1 x 6 ) + ( 7 x 6 / 2 x 1 )
= (525 + 210 + 21)
= 756.