A) \[\frac{1}{6}\]
B) \[\frac{5}{6}\]
C) \[\frac{1}{3}\]
D) None of these
Correct Answer: A
Solution :
Let \[{{E}_{i}}\] denote the event that the \[{{i}^{th}}\] object goes to the \[{{i}^{th}}\] place, we have \[P({{E}_{i}})=\frac{(n-1)\,!}{n\,!}=\frac{1}{n},\forall \,\,i\] and \[P({{E}_{1}}\cap {{E}_{j}}\cap {{E}_{l}})=\frac{(n-3)\,\,!}{n\,\,!}\] for \[i<j<k\] Since we can choose 3 places out of n in \[{}^{n}{{C}_{3}}\] ways. The probability of the required event is \[{}^{n}{{C}_{3}}.\frac{(n-3)\,!}{n\,!}=\frac{1}{6}\].
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