The number of functions f : {1, 2, 3, 4} righ | vedclass

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(D) Given the relation $f(n) + \frac{1}{n} f(n+1) = 1$,we can write $f(n+1) = n(1 - f(n))$.For $n=3$: $f(4) = 3(1 - f(3))$. Since $f(4) \in \mathbb{Z}

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$2$

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(D) Given the relation $f(n) + \frac{1}{n} f(n+1) = 1$,we can write $f(n+1) = n(1 - f(n))$.For $n=3$: $f(4) = 3(1 - f(3))$. Since $f(4) \in \mathbb{Z}$ and $|f(4)| \leq 8$,$f(4)$ must be a multiple of $3$ such that $|f(4)| \leq 8$. Possible values for $f(4)$ are $\{-6, -3, 0, 3, 6\}$.For $n=2$: $f(3) = 2(1 - f(2))$. Thus $f(3)$ must be an even integer such that $|f(3)| \leq 8$. Possible values for $f(3)$ are $\{-8, -6, -4, -2, 0, 2, 4, 6, 8\}$.For $n=1$: $f(2) = 1(1 - f(1)) = 1 - f(1)$. Thus $f(2)$ can be any integer such that $|f(2)| \leq 8$.We check the constraints:$1$. $f(4) = 3(1 - f(3)) \implies f(3) = 1 - \frac{f(4)}{3}$. For $f(3)$ to be an even integer,$\frac{f(4)}{3}$ must be an odd integer. Thus $f(4) \in \{-3, 3\}$.$2$. If $f(4) = -3$,then $f(3) = 1 - (-1) = 2$. Then $f(2) = 1 - \frac{f(3)}{2} = 1 - 1 = 0$. Then $f(1) = 1 - f(2) = 1 - 0 = 1$. All values are in the range $[-8, 8]$.$3$. If $f(4) = 3$,then $f(3) = 1 - 1 = 0$. Then $f(2) = 1 - \frac{f(3)}{2} = 1 - 0 = 1$. Then $f(1) = 1 - f(2) = 1 - 1 = 0$. All values are in the range $[-8, 8]$.Thus,there are $2$ such functions.

The number of functions $f : \{1, 2, 3, 4\} \rightarrow \{ a \in \mathbb{Z} : |a| \leq 8 \}$ satisfying $f(n) + \frac{1}{n} f(n+1) = 1$ for all $n \in \{1, 2, 3\}$ is

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