Question

There are n seats round a table numbered 1,2,3,…….., n. The number of ways in which $m\left( \le n \right)$ persons can take seats is:
A. $^{n}{{P}_{m}}$
B. $^{n}{{C}_{m}}\times \left( m-1 \right)!$
C. $^{n-1}{{P}_{m-1}}$
D. $^{n}{{C}_{m}}\times m!$

Answer 1

Hint: The number of ways in which ‘m’ persons can take place will be equal to the number of ways of selecting m seats out of ‘n’ seats multiplied by the number of ways of arrangement of ‘m’ persons on ‘n’ seats.

Complete step-by-step answer:
According to the question, there are ‘n’ seats numbered 1,2,………,n. and we have to find the number of ways in which ‘m’ persons can take seats out of these ‘n’ seats. And it is given that $m\le n$ and there are no restrictions.
To find this, we will first find the number of ways of selecting ‘m’ seats out of ‘n’ seats and then consider the arrangement of ‘m’ seats.
We know that no. of ways of selecting ‘r’ objects out of ‘n’ objects ${{=}^{n}}{{C}_{r}}$ .
So, No. Of ways of selecting ‘m’ seats out of ‘n’ seats ${{=}^{n}}{{C}_{m}}$ ………………. (1)
We know that No. of ways of arrangements of ‘n’ objects on ‘n’ different places =n!
So, no. of ways of arrangement of ‘m’ persons on ‘m’ seats =m!...................... (2)
And no. of ways in which ‘m’ persons can take seats out of ‘n’ different seats $\begin{align}
  & =\left( \text{No}\text{. of ways of selection of }\!\!'\!\!\text{ m }\!\!'\!\!\text{ seats out of }\!\!'\!\!\text{ n }\!\!'\!\!\text{ seats} \right)\times \\
 & \left( \text{No}\text{. of ways of arrangement of }\!\!'\!\!\text{ m }\!\!'\!\!\text{ persons on }\!\!'\!\!\text{ m }\!\!'\!\!\text{ seats} \right) \\
\end{align}$
$^{n}{{C}_{m}}\times m!$ [Using equation (1) and (2)].
Thus, no of ways in which ‘m’ persons can take seats out of ‘n’ seats ${{=}^{n}}{{C}_{m}}\times m!$ .
We know $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ .
So, the required number of ways $=\dfrac{n!}{m!\left( n-m \right)!}\times m!$
$=\dfrac{n!}{\left( n-m \right)!}$ .
We know $^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$ , So, $\dfrac{n!}{\left( n-m \right)!}{{=}^{n}}{{P}_{m}}$ .
Hence, the required number of ways in $^{n}{{P}_{m}}$ and $^{n}{{P}_{m}}$can be written as $\left( ^{n}{{C}_{m}}\times m! \right)$ . So, option A and D are correct.

So, the correct answer is “Option A and D”.

Note: Students can make mistakes here by using the concept of circular permutation as it is given that there is a round table. But, here we can’t use circular permutation because the seats are marked which means the seats will be considered different from each other.