Question

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

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

Answer 1

The correct option is D None of these

Total number of possible permutations = $n!$

Let $(1,2,3,...,k)$ be taken as group G.

Therefore there are now $n-k+1$ elements left i.e., $[G,k+1,...,n]$

which can be arranged in $(n-k+1)!$

Therefore, the required probability = $\frac{(n-k+1)!}{n!}$