Given, $$f(1) + f(2) = f(3)$$

It means $$f(1),f(2)$$ and $$f(3)$$ are dependent on each other. But there is no condition on $$f(4)$$, so $$f(4)$$ can be $$f(4) = 1,2,3,4,5,6$$.

For $$f(1),f(2)$$ and we have to find how many functions possible which will satisfy the condition $$f(1) + f(2) = f(3)$$

Case 1 :

When $$f(3) = 2$$ then possible values of $$f(1)$$ and $$f(2)$$ which satisfy $$f(1) + f(2) = f(3)$$ is $$f(1) = 1$$ and $$f(2) = 1$$.

And $$f(4)$$ can be = 1, 2, 3, 4, 5, 6

$$\therefore$$ Total possible functions $$=1\times6=6$$

Case 2 :

When $$f(3) = 3$$ then possible values

(1) $$f(1) = 1$$ and $$f(2) = 2$$

(2) $$f(1) = 2$$ and $$f(2) = 1$$

And $$f(4)$$ can be = 1, 2, 3, 4, 5, 6.

$$\therefore$$ Total functions $$ = 2 \times 6 = 12$$

Case 3 :

When $$f(3) = 4$$ then

(1) $$f(1) = 1$$ and $$f(2) = 3$$

(2) $$f(1) = 2$$ and $$f(2) = 2$$

(3) $$f(1) = 3$$ and $$f(2) = 1$$

And $$f(4)$$ can be = 1, 2, 3, 4, 5, 6

$$\therefore$$ Total functions $$ = 3 \times 6 = 18$$

Case 4 :

When $$f(3) = 5$$ then

(1) $$f(1) = 1$$ and $$f(4) = 4$$

(2) $$f(1) = 2$$ and $$f(4) = 3$$

(3) $$f(1) = 3$$ and $$f(4) = 2$$

(4) $$f(1) = 4$$ and $$f(4) = 1$$

And $$f(4)$$ can be = 1, 2, 3, 4, 5 and 6

$$\therefore$$ Total functions $$ = 4 \times 6 = 24$$

Case 5 :

When $$f(3) = 6$$ then

(1) $$f(1) = 1$$ and $$f(2) = 5$$

(2) $$f(1) = 2$$ and $$f(2) = 4$$

(3) $$f(1) = 3$$ and $$f(2) = 3$$

(4) $$f(1) = 4$$ and $$f(2) = 2$$

(5) $$f(1) = 5$$ and $$f(2) = 1$$

And $$f(4)$$ can be = 1, 2, 3, 4, 5 and 6

$$\therefore$$ Total possible functions $$ = 5 \times 6 = 30$$

$$\therefore$$ Total functions from those 5 cases we get

$$ = 6 + 12 + 18 + 24 + 30$$

$$ = 90$$