The total number of functions, $$f: \{1, 2, 3, 4\} \to \{1, 2, 3, 4, 5, 6\}$$ such that $$f(1) + f(2) = f(3)$$, is equal to

We need to count the total number of functions $$ f: \{1, 2, 3, 4\} \to \{1, 2, 3, 4, 5, 6\} $$ such that $$ f(1) + f(2) = f(3) $$.

We need $$ f(1) + f(2) = f(3) $$, where $$ f(1), f(2) \in \{1, 2, 3, 4, 5, 6\} $$ and $$ f(3) \in \{1, 2, 3, 4, 5, 6\} $$.

So we need $$ 2 \le f(1) + f(2) \le 6 $$ (since both are at least 1, and the sum must be at most 6).

Count pairs $$ (f(1), f(2)) $$ for each value of $$ f(3) $$:

• $$ f(3) = 2 $$: $$ f(1) + f(2) = 2 $$: only $$ (1,1) $$ — 1 pair
• $$ f(3) = 3 $$: $$ f(1) + f(2) = 3 $$: $$ (1,2), (2,1) $$ — 2 pairs
• $$ f(3) = 4 $$: $$ f(1) + f(2) = 4 $$: $$ (1,3), (2,2), (3,1) $$ — 3 pairs
• $$ f(3) = 5 $$: $$ f(1) + f(2) = 5 $$: $$ (1,4), (2,3), (3,2), (4,1) $$ — 4 pairs
• $$ f(3) = 6 $$: $$ f(1) + f(2) = 6 $$: $$ (1,5), (2,4), (3,3), (4,2), (5,1) $$ — 5 pairs

Total valid triples = $$ 1 + 2 + 3 + 4 + 5 = 15 $$.

$$ f(4) $$ can be any value in $$ \{1, 2, 3, 4, 5, 6\} $$, so there are 6 choices.

$$\text{Total} = 15 \times 6 = 90$$

The total number of functions is 90, which corresponds to Option B.