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 places are still occupied. The probability that both the places neighbouring his car are empty is
a) (r-1)!/(n-1)!
b)(r-1)!(n-r)!/(n-1)!
c)(N-r)(n-r+1)/(n+1)(n+2)
d)(N-r)c2/(n-1)c2
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1 Answers
62N cars are in a row
out of these N-r were removed.
2 of his neighboring are removed
So among the N-3 cars, N-r-2 are removed
So no of ways is N-3CN-r-2
total no of ways in which N-r cars could have been removed is
N-1CN-r
Now divide.. I think you should get the answer :)
