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(B) Given $A = \{1, 2, 3, 4, 5, 6\}$.We are given the condition $f(m) + f(n) = 7$ whenever $m + n = 7$.The pairs $(m, n)$ such that $m + n = 7$ are $(

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$216$

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(B) Given $A = \{1, 2, 3, 4, 5, 6\}$.We are given the condition $f(m) + f(n) = 7$ whenever $m + n = 7$.The pairs $(m, n)$ such that $m + n = 7$ are $(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)$.This implies the following constraints:$f(1) + f(6) = 7$$f(2) + f(5) = 7$$f(3) + f(4) = 7$For each pair,say $(f(1), f(6))$,the possible values for $f(1)$ are $\{1, 2, 3, 4, 5, 6\}$.If $f(1) = 1$,then $f(6) = 6$.If $f(1) = 2$,then $f(6) = 5$.If $f(1) = 3$,then $f(6) = 4$.If $f(1) = 4$,then $f(6) = 3$.If $f(1) = 5$,then $f(6) = 2$.If $f(1) = 6$,then $f(6) = 1$.There are $6$ possible choices for each pair $(f(1), f(6)), (f(2), f(5)),$ and $(f(3), f(4))$.Since there are $3$ such independent pairs,the total number of functions is $6 	imes 6 	imes 6 = 6^3 = 216$.

Let $A = \{1, 2, 3, 4, 5, 6\}$. The number of functions $f: A \to A$ such that $f(m) + f(n) = 7$ whenever $m + n = 7$ is:

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