Probability of Specific Student Seating and Derangement of Remaining Students

To find the probability that student $S_1$ receives their originally assigned seat $R_1$, while none of the remaining four students ($S_2, S_3, S_4, S_5$) receive their originally assigned seats ($R_2, R_3, R_4, R_5$), we analyze the total number of arrangements and the favorable arrangements.

Step 1: Total Number of Possible Outcomes

The total number of ways to randomly assign 5 distinct seats to 5 distinct students is given by the total permutations of 5 items: $\text{Total outcomes} = 5! = 120$

Step 2: Number of Favorable Outcomes

For a arrangement to be favorable:

1. Student $S_1$ must be assigned seat $R_1$. There is only $1$ way to place $S_1$ in $R_1$.
2. The remaining 4 students ($S_2, S_3, S_4, S_5$) must be assigned the remaining 4 seats ($R_2, R_3, R_4, R_5$) such that no student receives their original seat. This is a classic derangement problem of $n = 4$ elements.

The number of derangements of $n$ objects, denoted as $D_n$, is given by the formula: $D_n = n! \sum_{k=0}^{n} \frac{(-1)^k}{k!}$

For $n = 4$: $D_4 = 4! \left( \frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} \right)$ $D_4 = 24 \left( 1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} \right)$ $D_4 = 24 \times \frac{12 - 4 + 1}{24} = 9$

Thus, the number of favorable seating arrangements is: $\text{Favorable outcomes} = 1 \times D_4 = 9$

Step 3: Probability Calculation

The required probability $P$ is the ratio of favorable outcomes to the total outcomes: $P = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{9}{120} = \frac{3}{40}$

Hence, the correct option is (A).