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

Given that the function $f : A \rightarrow B$ satisfies the condition $f(1) + f(2) = f(4) - 1$, where the set $A = \{1, 2, 3, 4, 5\}$ and the set $B = \{1, 2, 3, 4, 5, 6\}$. <br/><br/>We want to find out how many such functions exist. <br/><br/>First, observe that the condition $f(1) + f(2) = f(4) - 1$ can be rewritten as $f(1) + f(2) + 1 = f(4)$. So, the sum of $f(1), f(2),$ and 1 is equal to $f(4)$. Since $f(4)$ is a value in set B, it can take values from 1 to 6. <br/><br/>The maximum value of $f(1) + f(2) + 1$ can be $6 + 6 + 1 = 13$, but this is more than 6 (the maximum value of $f(4)$), so it's not possible. Thus, the maximum value of $f(4)$ in this case can be 6. <br/><br/>Let's now analyze the number of functions for each value of $f(4)$ from 3 to 6 (we start from 3 because $f(1)$ and $f(2)$ take values from set B and their minimum sum plus 1 is 3): <br/><br/>1. When $f(4) = 6$, then $f(1) + f(2) = 5$. The pairs $(f(1), f(2))$ that satisfy this equation are $(1, 4), (2, 3), (3, 2), (4, 1)$. For each of these 4 cases, the functions $f(3)$ and $f(5)$ can each take 6 values from set B, resulting in $4 \times 6 \times 6 = 144$ functions. <br/><br/>2. When $f(4) = 5$, then $f(1) + f(2) = 4$. The pairs $(f(1), f(2))$ that satisfy this equation are $(1, 3), (2, 2), (3, 1)$. For each of these 3 cases, the functions $f(3)$ and $f(5)$ can each take 6 values from set B, resulting in $3 \times 6 \times 6 = 108$ functions. <br/><br/>3. When $f(4) = 4$, then $f(1) + f(2) = 3$. The pairs $(f(1), f(2))$ that satisfy this equation are $(1, 2), (2, 1)$. For each of these 2 cases, the functions $f(3)$ and $f(5)$ can each take 6 values from set B, resulting in $2 \times 6 \times 6 = 72$ functions. <br/><br/>4. When $f(4) = 3$, then $f(1) + f(2) = 2$. The only pair $(f(1), f(2))$ that satisfies this equation is $(1, 1)$. For this case, the functions $f(3)$ and $f(5)$ can each take 6 values from set B, resulting in $1 \times 6 \times 6 = 36$ functions. <br/><br/>Adding the numbers of functions from all these cases, we get a total of $144 + 108 + 72 + 36 = 360$ functions from $A$ to $B$ that satisfy the given condition. <br/><br/>Therefore, the number of functions $f : A \rightarrow B$ satisfying the condition $f(1) + f(2) = f(4) - 1$ is 360.