The number of ways in which a committee of 3 women and 4 men be chosen from 8 women and 7 men. If Mr. X refuses to serve on thcommittee of Mr. Y is a member of the committee is

Options

A. 420
B. 840
C. 1540
D. 1400

Answer

To solve the problem of forming a committee of 3 women and 4 men from 8 women and 7 men, with the condition that Mr. X refuses to serve if Ms. Y is a member, we can follow these steps:

### Step 1: Calculate the total number of committees without restrictions.
We start by calculating the total number of ways to choose 3 women from 8 and 4 men from 7.

- The number of ways to choose 3 women from 8 is given by the combination formula \( \binom{n}{r} \):
\[ \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \]

- The number of ways to choose 4 men from 7 is:
\[ \binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \]

- Therefore, the total number of unrestricted committees is:
\[ \text{Total Committees} = \binom{8}{3} \times \binom{7}{4} = 56 \times 35 = 1960 \]

### Step 2: Calculate the number of committees where Mr. X and Ms. Y are both included.
Next, we need to find the number of committees where both Mr. X and Ms. Y are included. If Mr. X is included, we have to choose 3 more men from the remaining 6 men (since Mr. X is already chosen), and if Ms. Y is included, we have to choose 2 more women from the remaining 7 women.

- The number of ways to choose 2 women from the remaining 7 is:
\[ \binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7 \times 6}{2 \times 1} = 21 \]

- The number of ways to choose 3 men from the remaining 6 is:
\[ \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]

- Therefore, the total number of committees where both Mr. X and Ms. Y are included is:
\[ \text{Committees with X and Y} = \binom{7}{2} \times \binom{6}{3} = 21 \times 20 = 420 \]

### Step 3: Subtract the restricted committees from the total committees.
Now, we subtract the number of committees that include both Mr. X and Ms. Y from the total number of committees calculated in Step 1.

- The number of valid committees (where Mr. X and Ms. Y are not both included) is:
\[ \text{Valid Committees} = \text{Total Committees} - \text{Committees with X and Y} = 1960 - 420 = 1540 \]

### Final Answer:
Thus, the number of ways to form a committee of 3 women and 4 men, such that Mr. X does not serve on the committee if Ms. Y is a member, is **1540**.

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