The number of ways of choosing a committee of 4 women and 5 men from 10 women and 9 men, if Mr. A refuses to serve on the committee when Ms. B is a member of the committee, is

Options

A. 20580
B. 21000
C. 21580
D. all the above

Answer

To solve the problem of choosing a committee of 4 women and 5 men from a group of 10 women and 9 men, with the condition that Mr. A will not serve on the committee if Ms. B is a member, we can follow these steps:

### Step 1: Calculate the total number of ways to choose 4 women and 5 men without any restrictions.

We can use the combination formula \( nCr = \frac{n!}{r!(n-r)!} \).

- The number of ways to choose 4 women from 10 is given by:
\[ \binom{10}{4} = \frac{10!}{4!(10-4)!} = \frac{10!}{4!6!} = 210 \]

- The number of ways to choose 5 men from 9 is given by:
\[ \binom{9}{5} = \frac{9!}{5!(9-5)!} = \frac{9!}{5!4!} = 126 \]

- Therefore, the total number of ways to choose the committee without restrictions is:
\[ \text{Total Ways} = \binom{10}{4} \times \binom{9}{5} = 210 \times 126 = 26,460 \]

### Step 2: Calculate the number of ways when both Mr. A and Ms. B are selected.

If Mr. A is selected, then Ms. B cannot be selected. However, if we consider the scenario where Ms. B is selected, we need to calculate how many ways we can form the committee under this condition.

- If Ms. B is selected, we need to choose 3 more women from the remaining 9 women:
\[ \binom{9}{3} = \frac{9!}{3!(9-3)!} = 84 \]

- Since Mr. A is also selected, we need to choose 4 men from the remaining 8 men (excluding Mr. A):
\[ \binom{8}{4} = \frac{8!}{4!(8-4)!} = 70 \]

- Therefore, the number of ways to choose the committee with both Mr. A and Ms. B is:
\[ \text{Ways with A and B} = \binom{9}{3} \times \binom{8}{4} = 84 \times 70 = 5,880 \]

### Step 3: Calculate the number of ways where Mr. A and Ms. B do not come together.

To find the number of ways where Mr. A and Ms. B do not both serve on the committee, we subtract the number of ways where both are selected from the total number of ways calculated in Step 1.

- Thus, the number of favorable ways is:
\[ \text{Favorable Ways} = \text{Total Ways} - \text{Ways with A and B} = 26,460 - 5,880 = 20,580 \]

### Final Answer:
The number of ways to choose the committee such that Mr. A and Ms. B do not serve together is **20,580**.

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