A car is parked by owner amongst 25 cars in a row not at the two extremes. On his return he find that exactly 15 places are still occupied. The probability that both the neighbouring places are vacant is

Ans: 15/92

driver's car can be in any of (25–2)places(leaving 2 ends)

Remaining (15–1) cars can be in (25–1) remaiming places.

So,total no of cases =(25–2)! ×(25–1)C(15–1)

=23×24c14

Favourable no of cases:

Driver's car can be in any of the 25–2 places and 2 neighbouring places are empty.

Remaining 15–1 cars can be in 25–3 remaining places.

Favourable no. Of cases= (25–2)!×(25–3)c(15–1)

Required probability=22c14/24c14

=22!×10!/(8!×24!)=10×9/(24×23)=45/276 = 15/92