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

The number of ways of arranging n numbers is n!

n(S)=n!

E= In each order obtained, we must now arrange the digits 1, 2, ..., k as group and the n - k remaining digits. This can be done in (n - k + l)! ways.

n(E)=(n - k + l)!

 Therefore, the probability for the required event is$P\left(E\right)=\frac{\left(n-k+1\right)!}{\mathrm{n}!}$