If f: {1, 2, 3, 4} to {1, 2, 3, 4} is a funct | vedclass

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(B) We are given the condition $|f(\alpha) - \alpha| \leqslant 1$ for each $\alpha \in \{1, 2, 3, 4\}$.This implies $f(\alpha) \in \{\alpha - 1, \alph

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(B) We are given the condition $|f(\alpha) - \alpha| \leqslant 1$ for each $\alpha \in \{1, 2, 3, 4\}$.This implies $f(\alpha) \in \{\alpha - 1, \alpha, \alpha + 1\} \cap \{1, 2, 3, 4\}$.For $\alpha = 1$: $f(1) \in \{1-1, 1, 1+1\} \cap \{1, 2, 3, 4\} = \{0, 1, 2\} \cap \{1, 2, 3, 4\} = \{1, 2\}$. ($2$ choices)For $\alpha = 2$: $f(2) \in \{2-1, 2, 2+1\} \cap \{1, 2, 3, 4\} = \{1, 2, 3\}$. ($3$ choices)For $\alpha = 3$: $f(3) \in \{3-1, 3, 3+1\} \cap \{1, 2, 3, 4\} = \{2, 3, 4\}$. ($3$ choices)For $\alpha = 4$: $f(4) \in \{4-1, 4, 4+1\} \cap \{1, 2, 3, 4\} = \{3, 4\}$. ($2$ choices)Since the choices for each $\alpha$ are independent,the total number of such functions is $2 	imes 3 	imes 3 	imes 2 = 36$.

If $f: \{1, 2, 3, 4\} \to \{1, 2, 3, 4\}$ is a function such that $|f(\alpha) - \alpha| \leqslant 1$ for all $\alpha \in \{1, 2, 3, 4\}$,then the total number of such functions is:

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