Probability of No Two Consecutive Students Sitting Adjacent in Five Seats

To find the probability of the event $T_1 \cap T_2 \cap T_3 \cap T_4$, we need to determine the number of seat assignments where no two consecutive students $S_i$ and $S_{i+1}$ sit next to each other on the examination day.

Let $a_i = \text{pos}(S_i)$ be the position (seat number from $1$ to $5$) occupied by student $S_i$ for $i = 1, 2, 3, 4, 5$.

The event $T_1 \cap T_2 \cap T_3 \cap T_4$ requires that for every $i \in \{1, 2, 3, 4\}$: $|a_{i+1} - a_i| \ge 2$

Thus, $(a_1, a_2, a_3, a_4, a_5)$ is a permutation of $\{1, 2, 3, 4, 5\}$ such that the absolute difference between any two consecutive elements in the array is at least $2$.

The total number of possible seat assignments for $5$ students in $5$ seats is given by: $n(S) = 5! = 120$

Now, we count the number of favorable permutations $(a_1, a_2, a_3, a_4, a_5)$ by considering the possible choices for $a_1$:

1. Case $a_1 = 1$:

   • $a_2$ must be $\ge 3$.
   • If $a_2 = 3 \implies a_3 = 5 \implies a_4 = 2 \implies a_5 = 4 \quad \longrightarrow \quad (1, 3, 5, 2, 4)$
   • If $a_2 = 4 \implies a_3 = 2 \implies a_4 = 5 \implies a_5 = 3 \quad \longrightarrow \quad (1, 4, 2, 5, 3)$
   • (Total: 2 permutations)

2. Case $a_1 = 2$:

   • If $a_2 = 4 \implies a_3 = 1 \implies a_4 = 3 \implies a_5 = 5 \quad \longrightarrow \quad (2, 4, 1, 3, 5)$
   • If $a_2 = 4 \implies a_3 = 1 \implies a_4 = 5 \implies a_5 = 3 \quad \longrightarrow \quad (2, 4, 1, 5, 3)$
   • If $a_2 = 5 \implies a_3 = 3 \implies a_4 = 1 \implies a_5 = 4 \quad \longrightarrow \quad (2, 5, 3, 1, 4)$
   • (Total: 3 permutations)

3. Case $a_1 = 3$:

   • If $a_2 = 1 \implies a_3 = 4 \implies a_4 = 2 \implies a_5 = 5 \quad \longrightarrow \quad (3, 1, 4, 2, 5)$
   • If $a_2 = 1 \implies a_3 = 5 \implies a_4 = 2 \implies a_5 = 4 \quad \longrightarrow \quad (3, 1, 5, 2, 4)$
   • If $a_2 = 5 \implies a_3 = 2 \implies a_4 = 4 \implies a_5 = 1 \quad \longrightarrow \quad (3, 5, 2, 4, 1)$
   • If $a_2 = 5 \implies a_3 = 1 \implies a_4 = 4 \implies a_5 = 2 \quad \longrightarrow \quad (3, 5, 1, 4, 2)$
   • (Total: 4 permutations)

4. Case $a_1 = 4$:

   • By symmetry with $a_1 = 2$ (replacing $x$ with $6-x$):
     • $(4, 2, 5, 3, 1)$
     • $(4, 2, 5, 1, 3)$
     • $(4, 1, 3, 5, 2)$
   • (Total: 3 permutations)

5. Case $a_1 = 5$:

   • By symmetry with $a_1 = 1$ (replacing $x$ with $6-x$):
     • $(5, 3, 1, 4, 2)$
     • $(5, 2, 4, 1, 3)$
   • (Total: 2 permutations)

Summing the favorable cases: $n(E) = 2 + 3 + 4 + 3 + 2 = 14$

The required probability is: $P(T_1 \cap T_2 \cap T_3 \cap T_4) = \frac{n(E)}{n(S)} = \frac{14}{120} = \frac{7}{60}$

Correct Option: C