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Let $A=\{1,2,3,4,5,6\}$ number of functions $f$ from $A$ to $A$ such that $f(m)+f(n)=7$, whenever $m+n=7$ is
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The correct answer is:
216
We have, $A=\{1,2,3,4,5,6\}$
$$
f(m)+f(n)=7 \text { of } m+n=7
$$
Order of $(m, n)$ is $\{(1,6),(2,5),(3,4),(4,3),(5,2)$, $(6,1)\}$
$$
\begin{aligned}
& f(1)+f(6)=7, f(2)+f(5)=7, f(3)+f(4)=7 \\
& f(1)+f(6)=7 \text { has } 36 \times 2=72 \text { functions }
\end{aligned}
$$
Similar for $f(2)+f(5)=7$ and $f(3)+f(4)=72$
Total number of functions $=72+72+72=216$
$$
f(m)+f(n)=7 \text { of } m+n=7
$$
Order of $(m, n)$ is $\{(1,6),(2,5),(3,4),(4,3),(5,2)$, $(6,1)\}$
$$
\begin{aligned}
& f(1)+f(6)=7, f(2)+f(5)=7, f(3)+f(4)=7 \\
& f(1)+f(6)=7 \text { has } 36 \times 2=72 \text { functions }
\end{aligned}
$$
Similar for $f(2)+f(5)=7$ and $f(3)+f(4)=72$
Total number of functions $=72+72+72=216$
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