The number of ways of choosing a committee of two women and three men from five women and six men, if Mr.A refuses to serve on the committee if Mr. B is a member and Mr. B can only serve, if Miss C is the member of the committee is

Options

A. a. 608
B. b. 124
C. c. 168
D. d. 180

Answer

To solve the problem of choosing a committee of 2 women and 3 men from 5 women and 6 men with the given conditions, we will break it down into different cases based on the conditions provided.

### Step 1: Identify the conditions
1. Mr. A refuses to serve if Mr. B is a member.
2. Mr. B can only serve if Miss C is also a member.

### Step 2: Case 1 - Miss C is a member
If Miss C is included in the committee, then Mr. B must also be included, and Mr. A cannot be included.

- **Choosing Women:** We have 5 women, and since Miss C is already chosen, we need to choose 1 more woman from the remaining 4 women. This can be done in \( \binom{4}{1} \) ways.
- **Choosing Men:** We need to choose 2 more men from the remaining 5 men (since Mr. A is excluded). This can be done in \( \binom{5}{2} \) ways.

Calculating this:
\[ \text{Ways} = \binom{4}{1} \times \binom{5}{2} = 4 \times 10 = 40 \]

### Step 3: Case 2 - Mr. B is not a member
If Mr. B is not a member, then Mr. A can be included.

- **Choosing Women:** We can choose 2 women from the 5 women. This can be done in \( \binom{5}{2} \) ways.
- **Choosing Men:** We can choose 3 men from the remaining 5 men (since Mr. B is excluded). This can be done in \( \binom{5}{3} \) ways.

Calculating this:
\[ \text{Ways} = \binom{5}{2} \times \binom{5}{3} = 10 \times 10 = 100 \]

### Step 4: Case 3 - Miss C is not a member
If Miss C is not a member, Mr. B cannot be a member either (as per the conditions).

- **Choosing Women:** We can choose 2 women from the 4 remaining women (excluding Miss C). This can be done in \( \binom{4}{2} \) ways.
- **Choosing Men:** We can choose 3 men from the 5 remaining men (excluding both Mr. A and Mr. B). This can be done in \( \binom{5}{3} \) ways.

Calculating this:
\[ \text{Ways} = \binom{4}{2} \times \binom{5}{3} = 6 \times 10 = 60 \]

### Step 5: Total number of ways
Now, we add the number of ways from all the cases:
\[ \text{Total Ways} = 40 + 100 + 60 = 200 \]

### Final Answer
The total number of ways to choose the committee is **200**.

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