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Question: A car is parked among N cars standing in a row, but not at either end. On his return, the owner find...

A car is parked among N cars standing in a row, but not at either end. On his return, the owner finds that exactly 'r' of the N placed are still occupied. The probability that both the places neighbouring his car empty is -

A

B

(r−1)!(N−r)!(N−1)!\frac { ( \mathrm { r } - 1 ) ! ( \mathrm { N } - \mathrm { r } ) ! } { ( \mathrm { N } - 1 ) ! }

C

(N−r)(N−r−1)(N+1)(N+2)\frac { ( \mathrm { N } - \mathrm { r } ) ( \mathrm { N } - \mathrm { r } - 1 ) } { ( \mathrm { N } + 1 ) ( \mathrm { N } + 2 ) }

D

N−rC2N−1C2\frac { { } ^ { \mathrm { N } - \mathrm { r } } \mathrm { C } _ { 2 } } { { } ^ { \mathrm { N } - 1 } \mathrm { C } _ { 2 } }

Answer

N−rC2N−1C2\frac { { } ^ { \mathrm { N } - \mathrm { r } } \mathrm { C } _ { 2 } } { { } ^ { \mathrm { N } - 1 } \mathrm { C } _ { 2 } }

Explanation

Solution

There are r cars in N places. Total no. of selection of places out of N – 1 Places for r – 1 cars

N – 1Cr – 1 Ž

If neighbouring places are empty then r – 1 cars must be parked in N – 3 places so the favourable cases

N – 3Cr – 1 Ž (N−3)!(r−1)!(N−r−2)!\frac { ( \mathrm { N } - 3 ) ! } { ( \mathrm { r } - 1 ) ! ( \mathrm { N } - \mathrm { r } - 2 ) ! }

Required probability

= (N−3)!(r−1)!(N−r−2)!\frac { ( \mathrm { N } - 3 ) ! } { ( \mathrm { r } - 1 ) ! ( \mathrm { N } - \mathrm { r } - 2 ) ! } ×

= (N−r)(N−r−1)(N−1)(N−2)\frac { ( \mathrm { N } - \mathrm { r } ) ( \mathrm { N } - \mathrm { r } - 1 ) } { ( \mathrm { N } - 1 ) ( \mathrm { N } - 2 ) } = N−rC2N−1C2\frac { { } ^ { N - r } C _ { 2 } } { { } ^ { N - 1 } C _ { 2 } }