There are n different objects, 1, 2, 3, 4, ..., n, distributed at random in n places marked 1, 2, 3, ..., n. If p is the probability that at least three of the objects occupy places corresponding to their number, find 6p

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Solution

Let $E_{i}$ denote the event that the ith object goes to the ith place. We have
$P (E_{i} \cap E_{j} \cap E_{k}) = \frac{(n - 3)!}{n!}$ 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
$p =^{n} C_{3} \cdot \frac{(n - 3)!}{n!} = \frac{n!}{3! (n - 3)!} \cdot \frac{(n - 3)!}{n!} = \frac{1}{6}$
$\Rightarrow 6 p = 1$

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There are n different objects, 1, 2, 3, 4, ..., n, distributed at random in n places marked 1, 2, 3, ..., n. If p is the probability that at least three of the objects occupy places corresponding to their number, find 6p

Let Ei denote the event that the ith object goes to the ith place. We haveP(Ei∩Ej∩Ek)=(n−3)!n! for i<j<kSince we can choose 3 places out of n in nC3 ways, the probability of the required event isp=nC3⋅(n−3)!n!=n!3!(n−3)!⋅(n−3)!n!=16⇒6p=1