The numbers 1, 2, 3, ..., `n` are arrange in a random order. The probability that the digits 1, 2, 3, .., `k(k

Options

A. `1//n!`
B. `k!//n!`
C. `(n - k)!//n!`
D. `(n - k + 1)!//n!`

Answer

D

Explanation

The number of ways of arranging n numbers is n! 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 + 1)! Ways. Therefore, the probability for the required event is `(n - k + 1)!//n!`