WBJEE · Maths · Permutation Combination
The numbers \(1,2,3, \ldots . . \mathrm{n}\) are arrange in a random order. The probability that the digits \(1,2,3, \ldots ., k(k - A \(\frac{1}{n!}\)
- B \(\frac{k!}{n!}\)
- C \((n-k)!n!\)
- D \(\frac{(n-k+1)!}{n!}\)
- A \(\frac{1}{n!}\)
- B \(\frac{k!}{n!}\)
- C \((n-k)!n!\)
- D \(\frac{(n-k+1)!}{n!}\)
Answer & Solution
Correct Answer
(D) \(\frac{(n-k+1)!}{n!}\)
Step-by-step Solution
Detailed explanation
The number of ways of arranging n numbers is n! In each order obtained, we must now arrange the digits \(1,2, \ldots, \mathrm{k}\) as group and the \(\mathrm{n}-\mathrm{k}\) remaining digits. This can be done in \((\mathrm{n}-\mathrm{k}+1)\)! ways. Therefore, the probability for…
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