Let A and B be two finite sets having m and n | vedclass

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(B) The total number of mappings from a set $A$ with $m$ elements to a set $B$ with $n$ elements is given by $n^m$.The number of injective mappings (o

Vedclass · Accepted Answer

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

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(B) The total number of mappings from a set $A$ with $m$ elements to a set $B$ with $n$ elements is given by $n^m$.The number of injective mappings (one-to-one functions) from $A$ to $B$ is given by the number of permutations of $n$ elements taken $m$ at a time,which is $P(n, m) = \frac{n!}{(n - m)!}$.The probability of selecting an injection is the ratio of the number of injective mappings to the total number of mappings:$P = \frac{\frac{n!}{(n - m)!}}{n^m} = \frac{n!}{(n - m)! n^m}$.

Let $A$ and $B$ be two finite sets having $m$ and $n$ elements respectively such that $m \le n.$ $A$ mapping is selected at random from the set of all mappings from $A$ to $B$. The probability that the mapping selected is an injection is

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