Question

Let $A=\{1,2,3,....,10\}$ and $f:A\to A$ be defined as $f\left(k\right)=\{\begin{array}{cc}k+1& \text{if}k\text{is odd}\\ k& \text{if}k\text{is even}\end{array}.$ Then the number of possible functions $g:A\to A$ such that $gof=f$ is:

• A. ${10}^5$
• B. ${}^{10}{C}_5$
• C. $5^5$
• D. $5!$

Answer 1

The correct option is A ${10}^5$

$g\left(f\right(x\left)\right)=f\left(x\right)$
$\Rightarrow g\left(x\right)=x$, when $x$ is even.
$5$ elements in $A$ can be mapped to any $10.$
So, ${10}^5\times 1={10}^5$