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# Let $A = {\text{ }}\left\{ {1,2,3,4,5} \right\}$ and $f:{\text{ }}A \to A$ ${\text{ }}f\left( i \right)\; \ne i\forall i \in A$ then number of such function f are

Last updated date: 20th Jun 2024
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Hint: To solve this question, we will check the possible choices of ‘i’ as ${\text{ }}f\left( i \right)\; \ne i\forall i \in A$. So, we have four possible choices for ‘i’ here, because here we have n - 1 = 4 choices for $f{\text{ }}:{\text{ }}\left\{ {1,2,3,4, \ldots .n} \right\}$ $\to \left\{ {1,2,3,4 \ldots .n} \right\}$ if $f\left( i \right) \ne {\text{i,}}$ and to find the number of such function, we will use the formula ${\left( {n - 1} \right)^n}$ for such functions. Thus, using the formula, we will get that there are ${4^5}$ such functions for f. Hence, after solving this we will get our required answer.

We have been given a set,$A = {\text{ }}\left\{ {1,2,3,4,5} \right\}$ and ${\text{ }}f:{\text{ }}A \to A{\text{ }}$ be an into function such that ${\text{ }}f\left( i \right)\; \ne i\forall i \in A$. We need to find all such functions of f.
So, for each i, there are only four choices are available because we know that there are $n{\text{ }} - {\text{ }}1$ choices for $f{\text{ }}:{\text{ }}\left\{ {1,2,3,4, \ldots .n} \right\}$ $\to \left\{ {1,2,3,4 \ldots .n} \right\}$ if $f\left( i \right) \ne {\text{i,}}$ and therefore, there are ${\left( {n - 1} \right)^n}$ such functions are there.
$n{\text{ }} = {\text{ }}5{\text{ }} \Rightarrow {\text{ (}}n{\text{ }}-{\text{ }}1) = {\text{ }}4$
As mentioned earlier, we will have ${\left( {n - 1} \right)^n}$ functions for f.
Therefore, ${4^5}$ such functions, i.e.,$4 \times 4 \times 4 \times 4 \times 4{\text{ }} = 1024$
Thus, $1024$ such functions are there.