Question

A car is parked among N car standing in a row, but not at either end. On his return, the owner finds that exactly ${}^{\prime }{r}^{\prime }$ of the N places are still occupied.The probability that the places neigbouring his car are empty is

• A. $\frac{(r-1{)}^{\prime }}{(N-1{)}^{\prime }}$
• B. $\frac{(r-1{)}^{\prime }(N-r{)}^{\prime }}{(N-1{)}^{\prime }}$
• C. $\frac{(N-r)(N-r-1)}{(N+1)(N+2)}$
• D. $\frac{{}^{N-r}{C}_2}{{}^{N-1}{C}_2}$

Answer 1

The correct option is D $\frac{{}^{N-r}{C}_2}{{}^{N-1}{C}_2}$

Out of his return, the owner finds that exactly r of the N places are still filled. This includes the space occupied by car as well..

Therefore, it means the remaining (r-1) cars are parked in (N-1) places.

Total number of ways = ${}^{N-1}{C}_{r-1}$

Since, the neighboring place to car are empty, it means (r-1) cars are parked among (N-3) places.

Number of favorable ways = ${}^{N-3}{C}_{r-1}$

Hence, the required probability, $P=\frac{\text{no. of favorable ways}}{\text{Total no. of ways}}$

$⟹=\frac{{}^{N-3}{C}_{r-1}}{{}^{N-1}{C}_{r-1}}⟹=\frac{(N-3)!}{(r-1)!(N-3-r+1)!}\times \frac{(r-1)!(N-1-r+1)!}{(N-1)!}⟹\frac{(N-3)!(N-r)!}{(N-1)!(N-r-2)!}=\frac{(N-1)(N-r-1)}{(N-1)(N-2)}=\frac{{}^{N-r}{C}_2}{{}^{N-1}{C}_2}$