The numbers 1, 2, 3,...,n are arranged in a random order. The probability that the digits 1, 2, 3,..., k (n>k) appear as neighbors is
Answer & explanation
Correct answer: option 2
The numbers 1, 2, 3,...,n can be arranged in a row in n! ways.
The total number of ways in which the digits 1, 2, 3,..., k (k <n) occur together is $k! (n-k + 1)!$
Hence, required probability =$\frac{k!(n-k+1)!}{n!}=\frac{n-k+1}{^nC_k}$