f(1)+f(2)=3−f(3) A={0,1,2,3,4,5,6,7} f:A→A So, f(1)+f(2)+f(3)=3 0+1+2=3 is the only possibility. So, f(0) can be either 0 or 1 or 2 . Similarly, f(1) and f(2) can be 0,1 and 2 . and
{3,4,5,6,7}
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⟶{3,4,5,6,7} They have 5 ! choices. And
{0,1,2}
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They have 3! choices. Number of bijective functions =3!×5!=720