The number of ways of choosing a committee from four men and six women so that the committee includes atleast two men and exactly twice as many women as men is
To form a committee with at least two men and exactly twice as many women as men, we can consider two cases: Case 1: The committee has 2 men and 4 women. There are ‌4C2 ways to choose 2 men from the 4 available, and ‌6C4 ways to choose 4 women from the 6 available. Therefore, there are ‌4C2⋅‌6C4 ways to form such a committee. Case 2: The committee has 3 men and 6 women. There are ‌4C3 ways to choose 3 men from the 4 available, and ‌6C6 ways to choose all 6 women. Therefore, there are ‌4C3⋅‌6C6 ways to form such a committee. The total number of ways to form a committee with at least two men and exactly twice as many women as men is the sum of the two cases: ‌4C2⋅‌6C4+‌4C3⋅‌6C6=90+4=94.