Question:
Let $A=\{0,1,2,3,4,5,6,7\} .$ Then the number of bijective functions $f: \mathrm{A} \rightarrow \mathrm{A}$ such that $f(1)+f(2)=3-f(3)$ is equal to
Solution:
$f(1)+f(2)=3-f(3)$
$\Rightarrow \mathrm{f}(1)+\mathrm{f}(2)=3+\mathrm{f}(3)=3$
The only possibility is : $0+1+2=3$
$\Rightarrow$ Elements $1,2,3$ in the domain can be mapped
with $0,1,2$ only.
So number of bijective functions.
$=\lfloor 3 \times\lfloor 5=720$
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