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(C) The total number of functions from a set $A$ with $n$ elements to itself is given by $n^n$.An injection (one-to-one function) from a finite set $A

Vedclass · Accepted Answer

$\frac{(n-1)!}{n^{n-1}}$

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(C) The total number of functions from a set $A$ with $n$ elements to itself is given by $n^n$.An injection (one-to-one function) from a finite set $A$ to itself is necessarily a bijection (one-to-one and onto function).The total number of bijections (permutations) of a set with $n$ elements is $n!$.Therefore,the probability that a randomly selected mapping is an injection is the ratio of the number of bijections to the total number of functions:$P = \frac{n!}{n^n} = \frac{n 	imes (n-1)!}{n 	imes n^{n-1}} = \frac{(n-1)!}{n^{n-1}}$.

$A$ mapping is selected at random from the set of all the mappings of the set $A = \{1, 2, ..., n\}$ into itself. The probability that the mapping selected is an injection is

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