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

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Question

(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

Vedclass · Accepted Answer

$\frac{1}{6}$

Vedclass · Answer

(a) Let ${E_i}$ denote the event that the ${i^{th}}$ object goes to the ${i^{th}}$ place,

we have $P({E_i}) = \frac{{(n - 1)\,!}}{{n\,!}} = \frac{1}{n},\forall \,\,i$

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 ${}^n{C_3}.\frac{{(n - 3)\,!}}{{n\,!}} = \frac{1}{6}$.

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

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