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

Let $$A = \{1,2,3,4,5\}$$, $$B = \{1,2,3,4,5,6\}$$. Find the number of functions $$f: A \to B$$ with $$f(1) + f(2) = f(4) - 1$$.

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

Let $$f(4) = k$$, so $$f(1) + f(2) = k - 1$$.

Since $$f(1), f(2) \geq 1$$, we need $$k - 1 \geq 2$$, i.e., $$k \geq 3$$.

Since $$f(1), f(2) \leq 6$$, we need $$k - 1 \leq 12$$, always satisfied.

$$k = 3$$: $$f(1) + f(2) = 2$$. Only $$(1,1)$$ → 1 pair

$$k = 4$$: $$f(1) + f(2) = 3$$. Pairs: $$(1,2), (2,1)$$ → 2 pairs

$$k = 5$$: $$f(1) + f(2) = 4$$. Pairs: $$(1,3), (2,2), (3,1)$$ → 3 pairs

$$k = 6$$: $$f(1) + f(2) = 5$$. Pairs: $$(1,4), (2,3), (3,2), (4,1)$$ → 4 pairs

Total constrained combinations: $$1 + 2 + 3 + 4 = 10$$

$$f(3)$$ and $$f(5)$$ are unconstrained, each with 6 choices.

Free choices: $$6 \times 6 = 36$$

$$ \text{Total} = 10 \times 36 = 360 $$

The number of functions is 360.