Q41 / 160 · AP EAMCET 2021 (25 Aug Shift 1) · Mathematics

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

  • A
    525
  • Correct answer
    B
    216
  • C
    200
  • D
    729

Solution

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 $=72+72+72=216$

Hence, the correct answer is option: 'B'.

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