Question

The numbers $1,2,\mathrm{3....},n$ are arranged in random order. The probability that the digits $1,2,\mathrm{3.....},k(k<n)$ appear 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

The correct option is D $\frac{(n-k+1)!}{n!}$

The number of ways of arranging $n$ numbers is $n!$
In each order obtained, we must now arrange the digits $1,\mathrm{2...}k$ 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 $\frac{(n-k+1)!}{n!}$