There are n stations between two cities A and B. A train is to stop at three of these n stations. What is the probaility that no two of these three stations are consecutive ?

Options

A. `(n-3)/(n(n-1))`
B. `((n-3)(n-4))/((n-1)(n-2))`
C. `(n-4)/(n(n-1))`
D. `((n-3)(n-4))/(n(n-1))`

Answer

D

Explanation

The total number of ways of choosing 3 stations out of n stations is `.^(n)C_(3)`.
So, total number of ways `= .^(n)C_(3)`.
Let `x_(1)` be the number of stations before the first halting stations, `x_(2)` between first and second, `x_(3)` between second and third and `x_(4)` on the right of third station. Then, `x_(1) ge 0, x_(2) ge 1, x_(3) ge 1 " and " x_(4) ge 0` such that `x_(1)+x_(2)+x_(3)+x_(4)=n-3`
The total number of solutions of this equation is `.^(n-2)C_(3)`.
Hence, required probability `=(.^(n-2)C_(3))/(.^(n)C_(3))=((n-3)(n-4))/(n(n-1))`