The numbers 1, 2, 3, ..., `n`are arrange in a random order. The probability that the digits 1, 2, 3,.., `k(kA. `(1)/(n!)`B. `(k !)/(n !)`C. `((n-k)!)/(n !)`D. none of these

The numbers 1, 2, 3, ..., `n`are arrange in a random order. The probab

1.	The numbers 1, 2, 3, ..., `n`are arrange in a random order. The probability that the digits 1, 2, 3,.., `k(kA. `(1)/(n!)`B. `(k !)/(n !)`C. `((n-k)!)/(n !)`D. none of these
Answer» Correct Answer - D The number of ways of arranging n numbers in a row is n ! Considering digits 1,2,3,4,..,k as one digit, we have (n-k+1) digits which can be arranged in (n-k+1)! Ways. So, the total number of ways in the digits 1,2,3,..,k appear as neighbours in the same order is (n-k+1)!. Hence, required probability `=((n-k+1)!)/(n!)`

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The numbers 1, 2, 3, ..., `n`are arrange in a random order. The probability that the digits 1, 2, 3,.., `k(kA. `(1)/(n!)`B. `(k !)/(n !)`C. `((n-k)!)/(n !)`D. none of these

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