A committee of 4 member is to be selected from 7 men and 5 women . In how many ways can the committee be formed if it must contain at least 2 women

A committee of 4 members must contain at least 2 women. This means the possible compositions are:

• Exactly 2 women and 2 men
• Exactly 3 women and 1 man
• Exactly 4 women and 0 men

The formula for combinations is $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.

• Case 1: 2 women and 2 men Number of ways to choose 2 women from 5:

  $$\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10$$

  Number of ways to choose 2 men from 7:

  $$\binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7 \times 6}{2 \times 1} = 21$$

  Total ways for Case 1 = $10 \times 21 = 210$

• Case 2: 3 women and 1 man Number of ways to choose 3 women from 5:

  $$\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10$$

  Number of ways to choose 1 man from 7:

  $$\binom{7}{1} = \frac{7!}{1!(7-1)!} = 7$$

  Total ways for Case 2 = $10 \times 7 = 70$

• Case 3: 4 women and 0 men Number of ways to choose 4 women from 5:

  $$\binom{5}{4} = \frac{5!}{4!(5-4)!} = \frac{5 \times 4 \times 3 \times 2}{4 \times 3 \times 2 \times 1} = 5$$

  Number of ways to choose 0 men from 7:

  $$\binom{7}{0} = \frac{7!}{0!(7-0)!} = 1$$

  Total ways for Case 3 = $5 \times 1 = 5$

Total number of ways = Ways for Case 1 + Ways for Case 2 + Ways for Case 3

$$\text{Total ways} = 210 + 70 + 5 = 285$$

The committee can be formed in $\boxed{285}$ ways.