Question

A car is parked among $n$ cars standing in row, but not at either end. On his return, the owner find that exactly $r$ of the $n$ places are still occupied. The probability that both the places neighboring has car are empty is

• A. $\frac{{}^{n-r}{C}_2}{{}^{n-1}{C}_2}$
• B. $\frac{(n-r)(n-r-1)}{(n+1)(n+2)}$
• C. $\frac{(r-2)!}{(n-2)!}$
• D. None of these

Answer 1

The correct option is A $\frac{{}^{n-r}{C}_2}{{}^{n-1}{C}_2}$

The total number selection of places for $(r-1)$ cars $($except the owner's car$)$ out of $(n-1)$ places
${=}^{n-1}{C}_{r-1}=\frac{(n-1)!}{(r-1)!(n-r)!}$
If neighboring place are empty,

then $(r-1)$ cars must be parked in $(n-3)$ places
So, the favorable cases ${=}^{n-3}{C}_{r-1}=\frac{(n-3)!}{(r-1)!(n-r-2)!}$
$\therefore$ Required probability $=\frac{(n-3)!}{(r-1)!(n-r-2)!}\times \frac{(r-1)!(n-r)!}{(n-1)!}$

$=\frac{(n-r)(n-r-1)}{(n-1)(n-2)}=\frac{{}^{n-r}{C}_2}{{}^{n-1}{C}_2}$