Number of $4$-member committees from $3$ women and $5$ men that have at least $2$ women 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 $70$?
 A: You can select 2 or 3 from the 3 available women and select however many men from the 5 available to complete a committee of 4.
$${^3\mathrm C_2}~{^5\mathrm C_2}+{^3\mathrm C_3}~{^5\mathrm C_1} = 35$$
That is all.
A: 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. 
A: If you have 3 women and must choose atleast two then there are two possibilities 
1) either you can choose 2women and 2 men 
2) or you can choose 3 women and 1 man
No of ways to choose 2 women = 3C2 =3
No, of ways to choose 2 men = 5C2 = 10
No of ways to choose 3 women =1
No.of ways to choose 1 man = 5
Therefore the answer will be (3*10)+(1*5)=35
