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

Question

The numbers 1, 2, 3, ..., n are arrange in a random order. The probability that the digits 1, 2, 3, .., k(kA. (1)/(n!)B. (k !)/(n !)C. ((n-k)!)/(n !)` D. none of these Select the correct answer from above options

Answer 1

Correct Answer - D 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!