Q.A committee of $6$ is to be chosen from $10$ men and $7$ women so as to contain atleast $3$ men and $2$ women. In how many different ways can this be done if two particular women refuse to serve on the same committee.

The key idea is Permutations Without Repetition (combinations), with a restriction that two specific women cannot be together.

Step 1: Total valid committees without the restriction

We need at least 3 men and 2 women from 10 men and 7 women. The possible splits (men, women) are:

• (3 men, 3 women)
• (4 men, 2 women)
• (5 men, 1 woman) — but this violates "at least 2 women", so discard.
• (4 men, 3 women) — wait, total must be 6, so only (3,3) and (4,2) are possible.

Check: (3 men, 3 women) gives $\binom{10}{3} \binom{7}{3}$, and (4 men, 2 women) gives $\binom{10}{4} \binom{7}{2}$.

Total without restriction:

Step 2: Subtract committees where the two particular women are together

Let the two women be A and B. Count committees where both A and B are selected. …

The problem asks for the number of ways to choose a committee of 6 people from 10 men and 7 women, with two constraints: the committee must have at least 3 men and at least 2 women, and two specific women (call them A and B) refuse to serve together.

This is a classic "at least" problem combined with a "restricted pairs" condition. The natural approach is to first count all committees satisfying the gender condition, then subtract those where the two particular women are both present.

1. Count all committees with at least 3 men and at least 2 women

Since the committee has exactly 6 members, the possible splits (men, women) that satisfy both "at least 3 men" and "at least 2 women" are:

• 3 men, 3 women
• 4 men, 2 women
• 5 men, 1 woman — but this fails the "at least 2 women" condition, so it's invalid.
• 6 men, 0 women — also invalid.

So only two cases are possible: (3 men, 3 women) and (4 men, 2 women).

Case 1: 3 men and 3 women

Choose 3 men from 10: $\binom{10}{3}$ ways.

Choose 3 women from 7: $\binom{7}{3}$ ways.

Total for this case: $\binom{10}{3} \times \binom{7}{3}$.

Case 2: 4 men and 2 women

Choose 4 men from 10: $\binom{10}{4}$ ways.

Choose 2 women from 7: $\binom{7}{2}$ ways.

Total for this case: $\binom{10}{4} \times \binom{7}{2}$.

So the total number of committees satisfying the gender condition is:

Let's compute each:

• $\binom{10}{3} = 120$, $\binom{7}{3} = 35$ → product = $120 \times 35 = 4200$
• $\binom{10}{4} = 210$, $\binom{7}{2} = 21$ → product = $210 \times 21 = 4410$

Sum = $4200 + 4410 = 8610$.

So there are 8610 committees that meet the gender requirement, without any restriction on the two particular women.

2. Subtract committees where the two particular women are together

Now we need to remove those committees where both A and B are present. But we must also ensure that the gender condition still holds — we only subtract committees that satisfy both: (a) A and B are both in the committee, and (b) the committee has at least 3 men and at least 2 women.

If A and B are both chosen, then we have already selected 2 women. The remaining 4 members must be chosen from the remaining people: 10 men and 5 other women (since A and B are already taken). And the committee must still have at least 3 men and at least 2 women — but since we already have 2 women (A and B), the "at least 2 women" condition is automatically satisfied. So we only need to ensure at least 3 men in the final committee of 6.

Since A and B are fixed, we need to choose 4 more people from the pool of 10 men + 5 other women = 15 people. The possible splits for the remaining 4 are:

• 3 men, 1 woman → total men = 3, total women = 3 (A, B, +1) → valid
• 4 men, 0 women → total men = 4, total women = 2 (A and B) → valid …