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

none of these

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 $=\frac{(n - k +1) !}{n!}$