The number of ways in which a committee of 3 ladies and 4 gentlemen can be appointed from a group consisting of 8 ladies and 7 gentlemen, if Mrs. X refuses to serve in a committee if Mr. Yis its member, is

Options

A. 1960
B. 3240
C. 1540
D. None of the above

Answer

To solve the problem of forming a committee of 3 ladies and 4 gentlemen from a group of 8 ladies and 7 gentlemen, with the condition that Mrs. X refuses to serve if Mr. Y is a member, we can break down the solution into several steps.

### Step-by-Step Solution:

1. **Identify the Total Number of Ladies and Gentlemen**:
- We have 8 ladies and 7 gentlemen.

2. **Calculate the Total Combinations Without Restrictions**:
- First, we calculate the number of ways to choose 3 ladies from 8:
\[ \text{Number of ways to choose 3 ladies} = \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \]
- Next, we calculate the number of ways to choose 4 gentlemen from 7:
\[ \text{Number of ways to choose 4 gentlemen} = \binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \]

3. **Calculate Total Combinations Without Any Restrictions**:
- Multiply the combinations of ladies and gentlemen:
\[ \text{Total combinations} = \binom{8}{3} \times \binom{7}{4} = 56 \times 35 = 1960 \]

4. **Account for the Restriction (Mrs. X and Mr. Y)**:
- We need to consider the scenario where Mr. Y is chosen. If Mr. Y is selected, Mrs. X cannot be part of the committee.
- If Mr. Y is included, we have 6 gentlemen left (since Mr. Y is already chosen) and we need to select 3 more gentlemen from these 6.
- The number of ways to choose 3 gentlemen from the remaining 6 is:
\[ \text{Number of ways to choose 3 gentlemen} = \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]
- The number of ways to choose 3 ladies remains the same (since we are still choosing from 8 ladies):
\[ \text{Number of ways to choose 3 ladies} = 56 \]
- Therefore, the total combinations where Mr. Y is included:
\[ \text{Total combinations with Mr. Y} = 56 \times 20 = 1120 \]

5. **Final Calculation**:
- Subtract the combinations where Mr. Y is included from the total combinations:
\[ \text{Final combinations} = 1960 - 1120 = 840 \]

Thus, the number of ways to form the committee under the given conditions is **840**.