Let A = 0, 1, 2, 3, 4, 5, 6, 7 . Then the number of bijective...

Given the set $A = { 0, 1, 2, 3, 4, 5, 6, 7 } .$

And, also a function $f: A \to A$ 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 $= 3 ! \times 5 ! = 6 \times 120 = 720 .$