Question

There are five students ${{S}_{1}},{{S}_{2}},{{S}_{3}},{{S}_{4}},{{S}_{5}}$ in a music class and for them are five seats ${{R}_{1}},{{R}_{2}},{{R}_{3}},{{R}_{4}},{{R}_{5}}$arranged in a row, where initially the seat ${{R}_{i}}$ is allotted to student ${{S}_{i}}$ where i=1, 2, 3, 4 and 5. But on the examination day, five students are randomly allotted the five seats.
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 previously allotted seats to him/her is?
(a) $\dfrac{3}{40}$
(b) $\dfrac{1}{8}$
(c) $\dfrac{7}{40}$
(d) $\dfrac{1}{5}$

Answer 1

Hint: First, before proceeding for this, we must know that the question is asking the concept of derangement in which none of the students except ${{S}_{1}}$gets his/her original seat. Then, by using the formula for n elements to be deranged, we have $n!\left( 1-\dfrac{1}{1!}+\dfrac{1}{2!}-.......+\dfrac{1}{n!} \right)$. Then, we can also see that the total number of cases of the arrangement of 5 students on 5 seats is given by 5! And then by dividing we get the required probability.

Complete step-by-step solution:
In this question, we are supposed to find 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 previously allotted seats to him/her when there are five students ${{S}_{1}},{{S}_{2}},{{S}_{3}},{{S}_{4}},{{S}_{5}}$in a music class and for them are five seats ${{R}_{1}},{{R}_{2}},{{R}_{3}},{{R}_{4}},{{R}_{5}}$arranged in a row, where initially the seat ${{R}_{i}}$ is allotted to student ${{S}_{i}}$ where i=1, 2, 3, 4 and 5. But on the examination day, five students are randomly allotted the five seats.
So, before proceeding for this, we must know that the question is asking the concept of derangement in which none of the students except ${{S}_{1}}$gets his/her original seat.
Now, by using the formula for n elements to be deranged, we have:
$n!\left( 1-\dfrac{1}{1!}+\dfrac{1}{2!}-.......+\dfrac{1}{n!} \right)$
Here, we are clearly given that the remaining four students are the case for disarranging and we get the value of n as 4.
Then, by substituting the value of n as 4 in disarranging formula, we get:
$4!\left( 1-\dfrac{1}{1!}+\dfrac{1}{2!}-\dfrac{1}{3!}+\dfrac{1}{4!} \right)$
Then, after solving we get the favourable cases as:
$\begin{align}
  & 4\times 3\times 2\times 1\left( 1-\dfrac{1}{1}+\dfrac{1}{2\times 1}-\dfrac{1}{3\times 2\times 1}+\dfrac{1}{4\times 3\times 2\times 1} \right) \\
 & \Rightarrow 24\left( \dfrac{1}{2}-\dfrac{1}{6}+\dfrac{1}{24} \right) \\
 & \Rightarrow 24\left( \dfrac{12-4+1}{24} \right) \\
 & \Rightarrow 9 \\
\end{align}$
So, we get the favourable number of cases as 9.
Then, we can also see that the total number of cases of arrangement of 5 students on 5 seats is given by:
$\begin{align}
  & 5!=5\times 4\times 3\times 2\times 1 \\
 & \Rightarrow 120 \\
\end{align}$
So, we get the probability that, on the examination day, the student ${{S}_{1}}$gets the previously allotted seat ${{R}_{1}}$and NONE of the remaining students get the previously allotted seats to him/her by dividing favorable outcomes to total outcomes as:
$\dfrac{9}{120}=\dfrac{3}{40}$
So, we get the required probability at $\dfrac{3}{40}$.
Hence, option (a) is correct.

Note: Now, to solve these types of the questions we need to know some of the basics of factorial and arrangement beforehand to get the answer easily. So, to find the factorial of the number n, multiply the number n with (n-1) till it reaches 1. To understand let us find the factorial of 4.
$\begin{align}
  & 4!=4\times 3\times 2\times 1 \\
 & \Rightarrow 24 \\
\end{align}$