Question

A car is parked by an owner in a parking lot of $25$ cars in a row, including his car not at either end. On his return he finds that exactly $15$ placed are still occupied. The probability that both the neighboring places are empty is

• A. $\frac{91}{276}$
• B. $\frac{15}{184}$
• C. $\frac{15}{92}$
• D. None of these

Answer 1

Answer: (C)

$\frac{15}{92}$

Fixing $15$ cars in $25$ places, there are $14$ cars other than his own in $24$ cars.

Therefore, total number of ways = ${}^{24}{C}_{14}=\frac{24!}{14!(25-15)!}$

Now the $14$ cars must be parked in $(25-3=22)$ places, since neighboring places are empty.

Therefore, number of favorable ways = ${}^{22}{C}_{14}=\frac{22!}{14!(22-14)!}$

Hence the required probability = $\frac{\text{Favourable ways}}{\text{Total ways}}=\frac{22!}{14!8!}\times \frac{14!10!}{24!}=\frac{10\times 9}{23\times 24}=\frac{15}{92}$