Let A = {0, 1, 2, 3, 4, 5, 6, 7}. Then the number of bijective functions $f;A\to A$ such that $f\left(1\right) + f\left(2\right) = 3 – f\left(3\right)$ is equal to

$$\begin{gathered} \text{A=}\left\{0,1,2,3,4,5,6,7\right\}, \text{⨍} : \mathrm{A}\to \mathrm{A} \\ \text{⨍ is a bijective function} \\ \text{⨍ (1)+⨍ (2)=3-⨍ (3)} \\ \Rightarrow \text{⨍(1)+⨍(2)+⨍(3)=3} \\ \Rightarrow \text{⨍(1), ⨍(2), ⨍(3)∈}\left\{0,1,2\right\} \\ \text{⨍(0), ⨍(1), ⨍(2) can selected in 3! ways other } \\ \text{⨍(3), ⨍(4)…⨍(7) ……5! ways} \\ \therefore \mathrm{No}.\mathrm{of} \mathrm{objections}=3!\times 5!=6\times 120=720 \end{gathered}$$