Question:
Let $\mathrm{A}=\{1,2,3 \ldots \ldots, 10\}$ and $\mathrm{f}: \mathrm{A} \rightarrow \mathrm{A}$ be defined as
$f(k)=\left\{\begin{array}{cc}k+1 & \text { if } k \text { is odd } \\ k & \text { if } k \text { is even }\end{array}\right.$ Then the number of possible functions $g: A \rightarrow A$ such that gof $=f$ is :
Correct Option: 1
Solution:
$g(f(x))=f(x)$
$\Rightarrow g(x)=x$, when $x$ is even
5 elements in A can be mapped to any 10
So, $10^{5} \times 1=10^{5}$
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