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

Explanation:

Since, there are 15 cars in 25 places, total number of selection of places out of (25 -1) places out of (15 – 1) cars (excepting the owner’s car) is–

= 25−1C15−1=24C14

If neighbouring places are empty, then 14 cars must be parked in (25 – 3 = 22) places.

So, the favourable number of cases is-

= 25−3C15−1=22C14

Therefore, Required probability = = 22C1424C14

= 22!\ 14!×8! × 14!×10!\24!

= 10×9 \ 24×23=15\92