Q49Multiple choice
A committee of 6 is to be chosen from 10 men and 7 women so as to contain atleast 3 men and 2 women. In how many different ways can this be done if two particular women refuse to serve on the same committee. [Hint:At least 3 men and 2 women: The number of ways = 10C3 × 7C3 + 10C4 × 7C2. For 2 particular women to be always there: the number of ways = 10C4 + 10C3 × 5C1 . The total number of committees when two particular women are never together = Total – together.]