# Let A

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 :

1. (1) $10^{5}$

2. (2) ${ }^{10} \mathrm{C}_{5}$

3. (3) $5^{5}$

4. (4) $5 !$

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, $10^{5} \times 1=10^{5}$