Find the number of ways of arranging 15 students `A_1,A_2,........A_15` in a row such that (i) `A_2`, must be seated after `A_1 and A_2`, must come after `A_2` (ii) neither `A_2` nor `A_3` seated brfore `A_1`

Options

A. `(2!xx15!)/(3!)`
B. `(15!)/(3!)`
C. `2!15!`
D. None of these

Explanation

To solve the problem of arranging 15 students \( A_1, A_2, \ldots, A_{15} \) under the given conditions, we will break it down into two parts as specified in the question.

### Part (i): Arranging with conditions \( A_2 \) must be seated after \( A_1 \) and \( A_3 \) must come after \( A_2 \).

1. **Choose Seats for \( A_1, A_2, A_3 \)**:
We need to select 3 seats out of the 15 for \( A_1, A_2, \) and \( A_3 \). The number of ways to choose 3 seats from 15 is given by the combination formula:
\[ \binom{15}{3} \]

2. **Arrange \( A_1, A_2, A_3 \)**:
Since \( A_1 \) must come before \( A_2 \), and \( A_2 \) must come before \( A_3 \), there is only 1 way to arrange them in the chosen seats:
\[ 1 \text{ way} \]

3. **Arrange Remaining Students**:
After placing \( A_1, A_2, \) and \( A_3 \), we have 12 students left to arrange in the remaining 12 seats. The number of ways to arrange these 12 students is:
\[ 12! \]

4. **Total Arrangements for Part (i)**:
The total number of arrangements for part (i) is:
\[ \text{Total} = \binom{15}{3} \times 1 \times 12! = \binom{15}{3} \times 12! \]

### Part (ii): Neither \( A_2 \) nor \( A_3 \) seated before \( A_1 \).

1. **Choose Seats for \( A_1, A_2, A_3 \)**:
Again, we choose 3 seats from 15 for \( A_1, A_2, \) and \( A_3 \):
\[ \binom{15}{3} \]

2. **Arrange \( A_1, A_2, A_3 \)**:
Here, \( A_1 \) must be seated before both \( A_2 \) and \( A_3 \). There are 2 valid arrangements for \( A_2 \) and \( A_3 \) after \( A_1 \):
- \( A_1, A_2, A_3 \)
- \( A_1, A_3, A_2 \)

Thus, there are:
\[ 2 \text{ ways} \]

3. **Arrange Remaining Students**:
The remaining 12 students can still be arranged in the remaining 12 seats:
\[ 12! \]

4. **Total Arrangements for Part (ii)**:
The total number of arrangements for part (ii) is:
\[ \text{Total} = \binom{15}{3} \times 2 \times 12! \]

### Final Answers:
- For part (i): \( \binom{15}{3} \times 12! \)
- For part (ii): \( \binom{15}{3} \times 2 \times 12! \)