Let A = {1, 2, 3, ldots, 10} and f: A rightar | vedclass

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(A) Given $f(k) = \begin{cases} k+1, & 	ext{if } k 	ext{ is odd} \ k, & 	ext{if } k 	ext{ is even} \end{cases}$.We are given the condition $g(f(k

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$10^{5}$

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(A) Given $f(k) = \begin{cases} k+1, & 	ext{if } k 	ext{ is odd} \ k, & 	ext{if } k 	ext{ is even} \end{cases}$.We are given the condition $g(f(k)) = f(k)$ for all $k \in A$.If $k$ is even,$f(k) = k$. Thus,$g(f(k)) = g(k) = f(k) = k$. So,$g(k) = k$ for all even $k \in \{2, 4, 6, 8, 10\}$.If $k$ is odd,$f(k) = k+1$. Thus,$g(f(k)) = g(k+1) = f(k) = k+1$. Since $k+1$ is always even,this confirms that $g$ must map even numbers to themselves,which we already established.For odd $k \in \{1, 3, 5, 7, 9\}$,the condition $g(f(k)) = f(k)$ does not impose any restriction on the value of $g(k)$.Since $g: A ightarrow A$,for each of the $5$ odd values of $k$,$g(k)$ can be any of the $10$ elements in $A$.Therefore,the number of such functions $g$ is $10 	imes 10 	imes 10 	imes 10 	imes 10 = 10^{5}$.

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

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