JEE Main 2021 — Functions Question with Solution

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

And, also a function $f:A\to A$ satisfying $f\left(1\right)+f\left(2\right)=3-f\left(3\right)$

$\Rightarrow f\left(1\right)+f\left(2\right)+f\left(3\right)=3$

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 $=3!\times 5!=6\times 120=720.$