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The numbers 1, 2, 3, ..., n are arrange ...

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 (a)`1 n !` (b)`k !`/`n!` (c) `(n-k)!`/`n!` (d)`(n-k+1)!`/`n!`

A

`1//n!`

B

`k!//n!`

C

`(n - k)!//n!`

D

`(n - k + 1)!//n!`

Text Solution

Verified by Experts

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!`
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