Question

There are m seats in the first row of a theatre of which n are to be occupied. Number of ways of arranging n persons so that there should be at least two empty seats between any two persons

• A. ${{}^{m-2n+2}}_{C_n}n!$
• B. ${{}^{m-n+1}}_{C_n}$ (n - 2)!
• C. ${{}^{m-n-1}}_{P_n}$
• D. ${{}^{m-n}}_{P_n}n!$

Answer 1

Answer: (A)

${{}^{m-2n+2}}_{C_n}n!$

Given that $n$ out of $m$ seats are to be occupied. Since there are $n$ people there are $(n+1)$ spaces between them (including the starting and ending seats which may or may not be occupied).

Let $x_1,x_2,x_3,......,x_{n+1}$ represent the spaces such that $x_1,x_{n+1}\ge 0$ $\xi$ $x_i\ge 2$

Where $i\in [2,n]$ and

$\Rightarrow x_1+x_2+x_3+......+x_{n+1}+n=m$

$\Rightarrow$ Let $x_1+x=y_1,$ $x_{n+1}=y$ and $x_i=y_i-2$ for $i=2$ to $n.$

$\Rightarrow y_i\ge 0$ for $i=1$ to $n+1$ and $y_1+(y_2-2)+(y_3-2)+.....+(y_n-2)+y_{n+1}=m-n$

$\Rightarrow y_1+y_2+....+y_{n+1}=m-n+2(n-1)$

$\Rightarrow y_1+y_2+....+y_{n+1}=m-n-2,y_i\ge 0$

Total combinations is given by ${}^{n+r-1}{C}_{r-1}$

here, $N=m+n-2,r=n+1$

Total $={}^{m+n-2+(n-1)-1}{C}_{(n+1)-1}$

$={}^{m+2n-2}{C}_n$

Now, these $n$ people can be arranged in their seats by $n!$ ways.

Total arrangement $=n!\times {}^{m+2n-2}{C}_n$

Hence, the answer is $n!\times {}^{m+2n-2}{C}_n.$