Given the set A=0,1,2,3,4,5,6,7.
And, also a function f:A→A satisfying f(1)+f(2)=3−f(3)
⇒f(1)+f(2)+f(3)=3
Since, the range of the function is also A, hence the only possibility satisfying the given condition is: 0+1+2=3
We know that, the number of arrangements of n objects at n places is n!.
Since, the given function is bijective i.e. one-one and onto, hence, the elements 1,2,3 in the domain can be mapped with only 0,1,2 in the co-domain in 3! ways and the remaining 5 elements 0,4,5,6,7 in the domain can be mapped with any of the remaining 5 elements 3,4,5,6,7 in 5! ways.
So, the number of bijective functions are =3!×5!=6×120=720.