PARAGRAPH A There are five students `S_1,\ S_2,\ S_3,\ S_4` and `S_5` in a music class and for them there are five seats `R_1,\ R_2,\ R_3,\ R_4` and `R_5` arranged in a row, where initially the seat `R_i` is allotted to the student `S_i ,\ i=1,\ 2,\ 3,\ 4,\ 5` . But, on the examination day, the five students are randomly allotted five seats. (For Ques. No. 17 and 18) The probability that, on the examination day, the student `S_1` gets the previously allotted seat `R_1` , and NONE of the remaining students gets the seat previously allotted to him/her is `3/(40)` (b) `1/8` (c) `7/(40)` (d) `1/5`

Options

A. `(3)/(40)`
B. `(1)/(8)`
C. `(7)/(40)`
D. `(1)/(5)`

Answer

To solve the problem, we need to calculate the probability that student \( S_1 \) gets his originally allotted seat \( R_1 \) and none of the other students \( S_2, S_3, S_4, S_5 \) get their originally allotted seats \( R_2, R_3, R_4, R_5 \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have 5 students \( S_1, S_2, S_3, S_4, S_5 \) and 5 seats \( R_1, R_2, R_3, R_4, R_5 \).
- \( S_1 \) must sit in \( R_1 \).
- The remaining students \( S_2, S_3, S_4, S_5 \) must not sit in \( R_2, R_3, R_4, R_5 \) respectively.

2. **Fixing \( S_1 \) in \( R_1 \)**:
- Since \( S_1 \) is fixed in \( R_1 \), we now need to arrange the remaining 4 students \( S_2, S_3, S_4, S_5 \) in the remaining 4 seats \( R_2, R_3, R_4, R_5 \).

3. **Counting Derangements**:
- We need to find the number of ways to arrange \( S_2, S_3, S_4, S_5 \) such that none of them sit in their originally allotted seats. This is known as a derangement.
- The formula for the number of derangements \( !n \) of \( n \) items is given by:
\[ !n = n! \sum_{i=0}^{n} \frac{(-1)^i}{i!} \]
- For \( n = 4 \):
\[ !4 = 4! \left( \frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} \right) \]
\[ = 24 \left( 1 - 1 + 0.5 - \frac{1}{6} + \frac{1}{24} \right) \]
\[ = 24 \left( 0 + 0.5 - 0.1667 + 0.0417 \right) \]
\[ = 24 \left( 0.375 \right) = 9 \]

4. **Total Arrangements**:
- The total number of unrestricted arrangements of 5 students is \( 5! = 120 \).

5. **Calculating the Probability**:
- The probability that \( S_1 \) is in \( R_1 \) and none of the other students are in their respective seats is given by:
\[ P = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{!4}{5!} = \frac{9}{120} = \frac{3}{40} \]

6. **Final Answer**:
- The probability that \( S_1 \) gets \( R_1 \) and none of the others get their seats is \( \frac{3}{40} \).