Question

The numbers $1,2,3,...,n$ are arranged in random order. The probability that the digits $1,2,...,k(k<n)$ appears as neighbours in that order is

• A. $\frac{1}{n!}$
• B. $\frac{k!}{n!}$
• C. $\frac{(n-k)!}{n!}$
• D. None of these

Answer 1

The correct option is D None of these

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