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

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Question

(A) Let $E_i$ be the event that the $i^{th}$ object occupies the $i^{th}$ place.The total number of ways to arrange $n$ objects in $n$ places is $n!$.

Vedclass · Accepted Answer

$\frac{1}{6}$

Vedclass · Answer

(A) Let $E_i$ be the event that the $i^{th}$ object occupies the $i^{th}$ place.The total number of ways to arrange $n$ objects in $n$ places is $n!$.The number of ways to choose $3$ objects out of $n$ that occupy their corresponding places is given by $\binom{n}{3}$.The remaining $(n-3)$ objects can be arranged in $(n-3)!$ ways.Thus,the number of ways where at least $3$ objects are in their correct places is $\binom{n}{3} 	imes (n-3)!$.The probability is $\frac{\binom{n}{3} 	imes (n-3)!}{n!} = \frac{n!}{3!(n-3)!} 	imes \frac{(n-3)!}{n!} = \frac{1}{3!} = \frac{1}{6}$.

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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