Let $A=\left\{1,2,3,4,5\right\}$ and $B=\left\{1,2,3,4,5,6\right\}$. Then the number of functions $f:A\to B$ satisfying $f\left(1\right)+f\left(2\right)=f\left(4\right)-1$ is equal to........

Given,

 $f\left(1\right)+f\left(2\right)+1=f\left(4\right)$

$\Rightarrow f\left(1\right)+f\left(2\right)\le 5$ $\left\{\text{as} f\left(4\right)\le 6\right\}$ 

Now taking cases we get,

Case $\left(1\right), f\left(1\right)=1\Rightarrow f\left(2\right)=\left\{1,2,3,4\right\}\to 4 \text{mappings}$

Case $\left(2\right), f\left(1\right)=2\Rightarrow f\left(2\right)=\left\{1,2,3\right\}\to 3 \text{mappings}$

Case $\left(3\right), f\left(1\right)=3\Rightarrow f\left(2\right)=\left\{1,2\right\}\to 2 \text{mappings}$

Case $\left(4\right), f\left(1\right)=4\Rightarrow f\left(2\right)=\left\{1\right\}\to 1 \text{mappings}$

And $f\left(5\right) \& f\left(6\right)\to 6 \text{mapping each}$

So, total number of function will be $=\left(4+3+2+1\right)\times 6\times 6=360$