CUET Preparation Today
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 |
$\frac{(n-k)!}{n!}$ $\frac{n-k+1}{^nC_k}$ $\frac{n-k}{^nC_k}$ $\frac{k!}{n!}$ |
$\frac{n-k+1}{^nC_k}$ |
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 =$\frac{k!(n-k+1)!}{n!}=\frac{n-k+1}{^nC_k}$ |
