1.Let A={1,2,3,…,10}A = \{1, 2, 3, \ldots, 10\}A={1,2,3,…,10} and f:A→Af: A \to Af:A→A be defined as f(k)={k+1,if k is oddk,if k is evenf(k) = \begin{cases} k+1, & \text{if } k \text{ is odd}\\ k, & \text{if } k \text{ is even} \end{cases}f(k)={k+1,k,if k is oddif k is even Then the number of possible functions g:A→Ag: A \to Ag:A→A such that g∘f=fg\circ f = fg∘f=f is:Login to continueOnly logged in users canattempt or see the solution.
