Question

1. from a group of 5 women and 4 men a committee of 4 is to be selected. in how many ways can the committee have at least two women?

Explanation:

Step1: Define valid committee cases

We consider 3 valid cases: 2 women + 2 men, 3 women + 1 man, 4 women + 0 men. We use combinations $C(n,k)=\frac{n!}{k!(n-k)!}$ for selections.

Step2: Calculate 2 women + 2 men

Compute combinations for each group and multiply:
$C(5,2) \times C(4,2) = \frac{5!}{2!3!} \times \frac{4!}{2!2!} = 10 \times 6 = 60$

Step3: Calculate 3 women + 1 man

Compute combinations for each group and multiply:
$C(5,3) \times C(4,1) = \frac{5!}{3!2!} \times \frac{4!}{1!3!} = 10 \times 4 = 40$

Step4: Calculate 4 women + 0 men

Compute combination for women (men selection is 1 way):
$C(5,4) \times C(4,0) = \frac{5!}{4!1!} \times 1 = 5 \times 1 = 5$

Step5: Sum all valid cases

Add the results from the three cases:
$60 + 40 + 5$

Answer:

105