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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

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Last updated date: 20th Jun 2024
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Answer
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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.

Complete step-by-step 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.
Now, on putting the values, we get
\[n{\text{ }} = {\text{ }}5{\text{ }} \Rightarrow {\text{ (}}n{\text{ }}-{\text{ }}1) = {\text{ }}4\]
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.

Note: In the question, we have been given about an into function. Let us understand about into function, so if there are two sets A and B, then in into function there is at least one element of set B which is not connected with the element of set A. There is also a method to determine whether a function is onto or into, so if range is equals to codomain, then function is onto and if range is a proper subset of the codomain, then function is into.