JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Sets Relations and Functions

The condition rearranges to $f(1)+f(2)+f(3)=3$. A bijection on $\{0,1,\dots ,7\}$ takes distinct values, and the only three distinct members of that set adding to $3$ are $0$, $1$ and $2$. These can be attached to $1$, $2$ and $3$ in $3!=6$ ways, and the five remaining elements of the domain take the five remaining values $3,4,5,6,7$ in $5!=120$ ways. The number of such bijective functions is $6\times 120=720$.