A committee of 6 is chosen from 10 men and 7 women so as to contain at least 3 men and 2 women. In how many ways can this be done if two particular women refuse to serve on the same committee? a. 850 b. 8700 c. 7800 d. none of these

Explanation

To solve the problem of forming a committee of 6 people from 10 men and 7 women, ensuring that there are at least 3 men and 2 women, while also considering that two particular women refuse to serve on the same committee, we can break down the solution into clear steps.

### Step-by-Step Solution:

1. **Identify Cases**:
We need to consider two cases based on the composition of men and women in the committee:
- Case 1: 3 men and 3 women
- Case 2: 4 men and 2 women

2. **Calculate for Case 1 (3 Men, 3 Women)**:
- Choose 3 men from 10:
\[ \binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \]
- Choose 3 women from 7:
\[ \binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \]
- Total ways for Case 1:
\[ 120 \times 35 = 4200 \]

3. **Calculate for Case 2 (4 Men, 2 Women)**:
- Choose 4 men from 10:
\[ \binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210 \]
- Choose 2 women from 7:
\[ \binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21 \]
- Total ways for Case 2:
\[ 210 \times 21 = 4410 \]

4. **Total Ways Without Restrictions**:
- Add the total ways from both cases:
\[ 4200 + 4410 = 8610 \]

5. **Account for the Condition of Two Particular Women**:
- We need to subtract the cases where the two particular women (let's call them W1 and W2) are both on the committee.
- If W1 and W2 are both on the committee, we treat them as a single unit. Thus, we have 5 remaining women and still need to choose 4 more members (2 men and 2 women).

6. **Calculate Cases with W1 and W2 Together**:
- **Sub-case 1**: Choose 3 men and 1 additional woman (from the remaining 5):
\[ \binom{10}{3} \times \binom{5}{1} = 120 \times 5 = 600 \]
- **Sub-case 2**: Choose 4 men and no additional women:
\[ \binom{10}{4} \times \binom{5}{0} = 210 \times 1 = 210 \]
- Total ways with W1 and W2 together:
\[ 600 + 210 = 810 \]

7. **Final Calculation**:
- Subtract the cases where W1 and W2 are together from the total ways:
\[ 8610 - 810 = 7800 \]

### Conclusion:
The total number of ways to form the committee under the given conditions is **7800**.