There are n different objects 1, 2, 3, dots, | vedclass

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(A) Let $E_i$ denote the event that the $i^{th}$ object goes to the $i^{th}$ place.We have $P(E_i) = \frac{(n-1)!}{n!} = \frac{1}{n}$ for all $i$.The

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$\frac{1}{6}$

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(A) Let $E_i$ denote the event that the $i^{th}$ object goes to the $i^{th}$ place.We have $P(E_i) = \frac{(n-1)!}{n!} = \frac{1}{n}$ for all $i$.The probability that exactly $3$ specific objects occupy their correct places is $P(E_i \cap E_j \cap E_k) = \frac{(n-3)!}{n!}$ for $i Since we can choose $3$ places out of $n$ in $\binom{n}{3}$ ways,the probability that at least three objects occupy their corresponding places is given by $\binom{n}{3} 	imes \frac{(n-3)!}{n!} = \frac{n!}{3!(n-3)!} 	imes \frac{(n-3)!}{n!} = \frac{1}{3!} = \frac{1}{6}$.

There are $n$ different objects $1, 2, 3, \dots, n$ distributed at random in $n$ places marked $1, 2, 3, \dots, n$. The probability that at least three of the objects occupy places corresponding to their number is

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