Question

A man parks his car among $n$ cars standing in a row, his car not being parked at an end. On his return he finds that exactly $m$ of the $n$ cars still there. What is the probability that both the cars parked on two sides of his car have left?

• A. $\frac{(n+m)(n-m+1)}{(n-1)(n-2)}$
• B. $\frac{(n-m)(n-m-1)}{(n-1)(n-2)}$
• C. $\frac{(n-m)(n-m+1)}{(n-1)(n-2)}$
• D. $\frac{(n+m)(n-m-1)}{(n-1)(n-2)}$

Answer 1

Answer: (B)

$\frac{(n-m)(n-m-1)}{(n-1)(n-2)}$

Clearly, his car is at one of the crosses(X).

$|XXX\cdots XXX|$

The number of the ways in which the remaining $(m-1)$ cars can take their places (excluding the car of the man)

$=n-1C_{m-1}$

{$\because$ There are $(n-1)$ places for the $(m-1)$ cars]

The number of ways in which the remaining $(m-1)$ cars can take places keeping the two places on two sides of his car $={}^{n-3}{C}_{m-1}$

$\therefore$ Required Probability

$=\frac{n\left(E\right)}{n\left(S\right)}=\frac{{}^{n-3}{C}_{m-1}}{{}^{n-1}{C}_{m-1}}$

$=\frac{(n-3)!}{(m-1)!(n-m-2)!}\times \frac{(m-1)!(n-m)!}{(n-1)!}$

$=\frac{(n-m)(n-m-1)}{(n-1)(n-2)}$