12 members of a committee are to sit down at random round a table. Probability that there are 3 members between the two particular members A and B, is

Options

A. `2/11`
B. `7/11`
C. `4/11`
D. `5/11`

Answer

To solve the problem of finding the probability that there are 3 members between two particular members A and B when 12 members of a committee are seated randomly around a table, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the Total Arrangements**:
When arranging \( n \) people around a round table, the number of ways to arrange them is given by \( (n-1)! \). For 12 members, the total arrangements are:
\[ \text{Total arrangements} = (12 - 1)! = 11! \]

2. **Identify the Favorable Cases**:
We need to find the arrangements where there are exactly 3 members between A and B. To visualize this, we can think of A and B as fixed points with 3 members between them.

3. **Fix A and B**:
Since A and B can be arranged in two ways (A can be to the left of B or to the right), we have:
\[ \text{Ways to arrange A and B} = 2 \]

4. **Select 3 Members Between A and B**:
We need to select 3 members from the remaining 10 members (since A and B are already chosen). The number of ways to choose 3 members from 10 is given by:
\[ \binom{10}{3} \]

5. **Arrange the 3 Members**:
The 3 members chosen can be arranged among themselves in:
\[ 3! \text{ ways} \]

6. **Arrange the Remaining Members**:
After fixing A, B, and the 3 members between them, there are 7 remaining members. These can be arranged in:
\[ 7! \]

7. **Calculate the Total Favorable Arrangements**:
Now, we combine all the arrangements:
\[ \text{Favorable arrangements} = 2 \times \binom{10}{3} \times 3! \times 7! \]

8. **Calculate the Probability**:
The probability \( P \) that there are 3 members between A and B is given by the ratio of favorable arrangements to total arrangements:
\[ P = \frac{2 \times \binom{10}{3} \times 3! \times 7!}{11!} \]

9. **Simplify the Expression**:
We know that \( 11! = 11 \times 10! \), and we can simplify:
\[ P = \frac{2 \times \frac{10!}{3! \times 7!} \times 6 \times 7!}{11 \times 10!} \]
\[ P = \frac{2 \times 10 \times 9 \times 8}{11 \times 6} \]
\[ P = \frac{2 \times 720}{66} = \frac{1440}{66} = \frac{240}{11} \]

10. **Final Result**:
The probability that there are 3 members between A and B is:
\[ P = \frac{2}{11} \]