Question

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 the five seats.

For $i=1,2,3,4,$ let $T_i$ denote the event that the students $S_i$ and $S_{i+1}$ do $NOT$ sit adjacent to each other on the day of the examination. Then, the probability of the event $T_1\cap T_2\cap T_3\cap T_4$ is

• A. $\frac{1}{15}$
• B. $\frac{1}{10}$
• C. $\frac{7}{60}$
• D. $\frac{1}{5}$

Answer 1

The correct option is C $\frac{7}{60}$

$T_1$: $S_1$ and $S_2$ cannot sit together,
$T_2$: $S_2$ and $S_3$ cannot sit together,
$T_3$: $S_3$ and $S_4$ cannot sit together,
$T_4$: $S_4$ and $S_5$ cannot sit together.
Now possible ways of seating arrangement,
Case 1: when $S_1$ is seating on $1^{st}$ place.
$\begin{array}{ccccc}{S}_1& \underset{–}{S_3}& \underset{–}{S_5}& \underset{–}{S_2}& \underset{–}{S_4}\\ & \underset{–}{S_4}& \underset{–}{S_2}& \underset{–}{S_5}& \underset{–}{S_3}\\ & \underset{–}{S_5}& \underset{–}{S_3}& \underset{–}{\times }& \underset{–}{\times }\\ & & \underset{–}{S_2}& \underset{–}{S_4}& \underset{–}{\times }\end{array}$
$\therefore$ 2 ways of seating arrangement.
Case 2: when $S_5$ is seating on $1^{st}$ place.
$\begin{array}{ccccc}{S}_5& \underset{–}{S_1}& \underset{–}{S_3}& \underset{–}{\times }& \underset{–}{\times }\\ & & \underset{–}{S_4}& \underset{–}{S_2}& \underset{–}{\times }\\ & \underset{–}{S_2}& \underset{–}{S_4}& \underset{–}{S_1}& \underset{–}{S_3}\\ & \underset{–}{S_3}& \underset{–}{S_1}& \underset{–}{S_4}& \underset{–}{S_2}\end{array}$
$\therefore$ 2 ways of seating arrangement.
Case 3: when $S_2$ is seating on $1^{st}$ place.
$\begin{array}{ccccc}{S}_2& \underset{–}{S_4}& \underset{–}{S_1}& \underset{–}{S_5}& \underset{–}{S_3}\\ & & & \underset{–}{S_3}& \underset{–}{S_5}\\ & \underset{–}{S_5}& \underset{–}{S_1}& \underset{–}{\times }& \underset{–}{\times }\\ & & \underset{–}{S_3}& \underset{–}{S_1}& \underset{–}{S_4}\end{array}$
$\therefore$ 3 ways of seating arrangement.
Case 4: when $S_4$ is seating on $1^{st}$ place.
$\begin{array}{ccccc}{S}_4& \underset{–}{S_1}& \underset{–}{S_3}& \underset{–}{S_5}& \underset{–}{S_2}\\ & & \underset{–}{S_5}& \underset{–}{S_3}& \underset{–}{\times }\\ & & & \underset{–}{S_2}& \underset{–}{\times }\\ & \underset{–}{S_2}& \underset{–}{S_5}& \underset{–}{S_3}& \underset{–}{S_1}\\ & & & \underset{–}{S_1}& \underset{–}{S_3}\end{array}$
$\therefore$ 3 ways of seating arrangement.
Case 5: when $S_3$ is seating on $1^{st}$ place.
$\begin{array}{ccccc}{S}_3& \underset{–}{S_1}& \underset{–}{S_4}& \underset{–}{S_2}& \underset{–}{S_5}\\ & & \underset{–}{S_5}& \underset{–}{S_2}& \underset{–}{S_4}\\ & \underset{–}{S_5}& \underset{–}{S_2}& \underset{–}{S_4}& \underset{–}{S_1}\\ & & \underset{–}{S_1}& \underset{–}{S_4}& \underset{–}{S_2}\end{array}$
$\therefore$ 4 ways of seating arrangement.
Number of seating arrangement when $(T_1\cap T_2\cap T_3\cap T_4)$ $=14$
Total arrangement $=5!=120$
Required probability
$P(T_1\cap T_2\cap T_3\cap T_4)=\frac{14}{120}=\frac{7}{60}$