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 the places neighboring his car are empty is `((r-1)!)/((N-1)!)` b. `((r-1)!(N-r)!)/((N-1)!)` c. `((N-r)(N-r-1))/((N-1)(N+2))` d. `(^(N-r)C_2)/(^(N-1)C_2)`

Options

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)C_(2))/(.^(N-1)C_(2))`

Answer

D

Explanation

Since there are r cars in N places, total number of selection of places out of N - 1 places for r - 1 cars (except the owner's car) is
`.^(N-1)C_(r-1)=((N-1)!)/((r-1)!(N-r)!)`
If neighboring places are empty, then r - 1 cars must be parked in N - 3 places, So, the favorable number of cases is
`.^(N-3)C_(r-1)=((N-3)!)/((r-1)!(N-r-2)!)`
Therefore, the required probability is
= `((N-3)!)/((r-1)!(N-r-2)!) xx ((r-1)!(N-r)!)/((N-1)!)`
`=((N-r)(N-r-1))/((N-1)(N-2)) = (.^(N-r)C_(2))/(.^(N-1)C_(2))`