Let A = {1, 2, 3, 4, 5} and B = {1, 2, 3, 4, | vedclass

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(A) Given the condition $f(1) + f(2) = f(4) - 1$,we can rewrite it as $f(1) + f(2) + 1 = f(4)$.Since the codomain $B = \{1, 2, 3, 4, 5, 6\}$,the maxim

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$360$

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(A) Given the condition $f(1) + f(2) = f(4) - 1$,we can rewrite it as $f(1) + f(2) + 1 = f(4)$.Since the codomain $B = \{1, 2, 3, 4, 5, 6\}$,the maximum value of $f(4)$ is $6$.Therefore,$f(1) + f(2) + 1 \leq 6$,which implies $f(1) + f(2) \leq 5$.We analyze the possible values for $f(1)$ and $f(2)$ where $f(1), f(2) \in \{1, 2, 3, 4, 5, 6\}$:Case $(i)$: If $f(1) = 1$,then $f(2) \in \{1, 2, 3, 4\}$ (since $1 + f(2) \leq 4$,$f(2) \leq 3$ is not correct,let's re-evaluate: $f(1)+f(2) \leq 5$. If $f(1)=1$,$f(2) \in \{1, 2, 3, 4\}$,total $4$ pairs).Case $(ii)$: If $f(1) = 2$,then $f(2) \in \{1, 2, 3\}$,total $3$ pairs.Case $(iii)$: If $f(1) = 3$,then $f(2) \in \{1, 2\}$,total $2$ pairs.Case $(iv)$: If $f(1) = 4$,then $f(2) = 1$,total $1$ pair.Total pairs for $(f(1), f(2))$ is $4 + 3 + 2 + 1 = 10$.For each pair,$f(4)$ is uniquely determined as $f(1) + f(2) + 1$.Since $f(3)$ and $f(5)$ can take any value from $B$ (which has $6$ elements),there are $6 	imes 6 = 36$ ways to choose $f(3)$ and $f(5)$.Total number of functions $= 10 	imes 6 	imes 6 = 360$.

Let $A = \{1, 2, 3, 4, 5\}$ and $B = \{1, 2, 3, 4, 5, 6\}$. Then the number of functions $f: A \rightarrow B$ satisfying $f(1) + f(2) = f(4) - 1$ is equal to

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