The numbers 1, 2, 3, ldots, n are arranged in | vedclass

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(D) The total number of ways to arrange $n$ distinct numbers is $n!$. To ensure the digits $1, 2, 3, \ldots, k$ appear as a single block in that speci

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$\frac{(n-k+1)!}{n!}$

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(D) The total number of ways to arrange $n$ distinct numbers is $n!$. To ensure the digits $1, 2, 3, \ldots, k$ appear as a single block in that specific order, we treat this block as a single entity. Now, we have $(n - k)$ remaining digits plus the one block, totaling $(n - k + 1)$ entities to arrange. The number of ways to arrange these $(n - k + 1)$ entities is $(n - k + 1)!$. Therefore, the probability of the required event is $\frac{(n - k + 1)!}{n!}$.

The numbers $1, 2, 3, \ldots, n$ are arranged in a random order. The probability that the digits $1, 2, 3, \ldots, k$ appear as a block in that order is:

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