The numbers 1, 2, 3, ..., `n` are arrange in a random order. The probability that the digits 1, 2, 3, .., `k( k < n)` appear as neighbours in that order is

Options

A. `(1)/(n!)`
B. `(k !)/(n !)`
C. `((n-k)!)/(n !)`
D. none of these

Answer

D

Explanation

The number of ways of arranging n numbers in a row is n !
Considering digits 1,2,3,4,..,k as one digit, we have (n-k+1) digits which can be arranged in (n-k+1)! Ways.
So, the total number of ways in the digits 1,2,3,..,k appear as neighbours in the same order is (n-k+1)!.
Hence, required probability `=((n-k+1)!)/(n!)`