Question

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

• A. $\frac{1}{n!}$
• B. $\frac{k!}{n!}$
• C. $\frac{(n - k)!}{n!}$
• D. $\frac{(n - k + 1 )!}{n!}$

Answer 1

Answer: (D)

Exhaustive number of cases = $n!$
Assuming the set of numbers {1, 2, 3, ..., k}
as one the favourable cases = (n - k + 1)!
$\therefore $ Probability $ = \frac{(n - k + 1)!}{n!}$