The United States Supreme Court consists of 3 women and 5 men. In how many ways can a 4-member committee be formed if each committee must have at least two women?.
I know that we have $^8C_4=70$ combinations.
I'm stuck on how many committees can be formed with at least two women.
Do I get the combination of committees that include all men and subtract with the $70$?
Best Answer
Check the following combinations:
$1.$ We choose $2$ women and $2$ men. The number of ways for doing so is: $$\binom {3}{2}\times \binom {5}{2} = 3\times 10 =30$$
$2.$ We choose $3$ women and $1$ man. The number of ways for doing so is: $$\binom {3}{3}\times \binom {5}{1} = 1\times 5 =5$$
$3.$ We can choose all $4$ as women. But there are only $3$ women available, so this is not possible.
Thus, the total number of ways to select equals: $30+5=35$ ways. Hope it helps.