Key Concepts and Formulas

• Functions: A function $f: A \rightarrow B$ assigns exactly one element from set $B$ to each element in set $A$.
• Multiplication Principle: If an event can occur in $m$ ways and another independent event can occur in $n$ ways, then both events can occur in $m \times n$ ways.
• Addition Principle: If an event can occur in $m$ ways and another mutually exclusive event can occur in $n$ ways, then either event can occur in $m + n$ ways.

Step-by-Step Solution

Step 1: Understand the Problem and Define the Constraints We are given a function $f$ with domain $A = \{1, 2, 3, 4, 5\}$ and codomain $B = \{1, 2, 3, 4, 5, 6\}$. The function must satisfy the condition $f(1) + f(2) = f(4) - 1$. We need to find the number of such functions. The condition can be rewritten as $f(4) = f(1) + f(2) + 1$. Since $f(x) \in B$ for all $x \in A$, we have $1 \le f(1) \le 6$, $1 \le f(2) \le 6$, and $1 \le f(4) \le 6$.

Step 2: Determine the Possible Values for $f(1), f(2),$ and $f(4)$ Using the rewritten condition $f(4) = f(1) + f(2) + 1$ and the bounds for $f(1)$ and $f(2)$: The minimum possible value for $f(1) + f(2)$ is $1 + 1 = 2$. The maximum possible value for $f(1) + f(2)$ is $6 + 6 = 12$. Now, let's find the range for $f(4)$: Minimum $f(4) = (f(1) + f(2))_{\min} + 1 = 2 + 1 = 3$. Maximum $f(4) = (f(1) + f(2))_{\max} + 1 = 12 + 1 = 13$. However, $f(4)$ must be in the codomain $B$, so $1 \le f(4) \le 6$. Combining these, the possible values for $f(4)$ are restricted to $\{3, 4, 5, 6\}$. We will perform a case analysis based on the value of $f(4)$.

Step 3: Case Analysis for $f(4)$ For each case of $f(4)$, we will find the number of pairs $(f(1), f(2))$ that satisfy the condition and then multiply by the number of choices for $f(3)$ and $f(5)$.

Case 1: $f(4) = 3$ The condition becomes $3 = f(1) + f(2) + 1$, which simplifies to $f(1) + f(2) = 2$. The only pair $(f(1), f(2))$ from $B \times B$ satisfying this is $(1, 1)$. So there is 1 way to choose $f(1)$ and $f(2)$. The value of $f(4)$ is fixed at 3 (1 way). For $f(3)$, there are 6 choices from $B$. For $f(5)$, there are 6 choices from $B$. Number of functions in this case = $1 \times 1 \times 6 \times 6 = 36$.

Case 2: $f(4) = 4$ The condition becomes $4 = f(1) + f(2) + 1$, which simplifies to $f(1) + f(2) = 3$. The pairs $(f(1), f(2))$ from $B \times B$ satisfying this are $(1, 2)$ and $(2, 1)$. So there are 2 ways to choose $f(1)$ and $f(2)$. The value of $f(4)$ is fixed at 4 (1 way). For $f(3)$, there are 6 choices from $B$. For $f(5)$, there are 6 choices from $B$. Number of functions in this case = $2 \times 1 \times 6 \times 6 = 72$.

Case 3: $f(4) = 5$ The condition becomes $5 = f(1) + f(2) + 1$, which simplifies to $f(1) + f(2) = 4$. The pairs $(f(1), f(2))$ from $B \times B$ satisfying this are $(1, 3)$, $(2, 2)$, and $(3, 1)$. So there are 3 ways to choose $f(1)$ and $f(2)$. The value of $f(4)$ is fixed at 5 (1 way). For $f(3)$, there are 6 choices from $B$. For $f(5)$, there are 6 choices from $B$. Number of functions in this case = $3 \times 1 \times 6 \times 6 = 108$.

Case 4: $f(4) = 6$ The condition becomes $6 = f(1) + f(2) + 1$, which simplifies to $f(1) + f(2) = 5$. The pairs $(f(1), f(2))$ from $B \times B$ satisfying this are $(1, 4)$, $(2, 3)$, $(3, 2)$, and $(4, 1)$. So there are 4 ways to choose $f(1)$ and $f(2)$. The value of $f(4)$ is fixed at 6 (1 way). For $f(3)$, there are 6 choices from $B$. For $f(5)$, there are 6 choices from $B$. Number of functions in this case = $4 \times 1 \times 6 \times 6 = 144$.

Step 4: Sum the Results from All Cases Since the cases for $f(4)$ are mutually exclusive, we use the Addition Principle to find the total number of functions. Total number of functions = (Functions for $f(4)=3$) + (Functions for $f(4)=4$) + (Functions for $f(4)=5$) + (Functions for $f(4)=6$) Total number of functions = $36 + 72 + 108 + 144 = 360$.

Common Mistakes & Tips

• Range of $f(x)$: Ensure that all function values $f(x)$ are within the codomain $B = \{1, 2, 3, 4, 5, 6\}$.
• Ordered Pairs: When finding pairs $(f(1), f(2))$ that sum to a certain value, remember that the order matters since $f(1)$ and $f(2)$ are values for distinct elements of the domain.
• Independent Choices: Do not forget to account for the choices for elements of the domain not involved in the constraint (here, $f(3)$ and $f(5)$). Each can map to any of the 6 elements in the codomain independently.

Summary We analyzed the given functional equation $f(1) + f(2) = f(4) - 1$ and determined the possible values for $f(4)$ to be $\{3, 4, 5, 6\}$ by considering the constraints imposed by the codomain. We then performed a case-by-case analysis for each possible value of $f(4)$, counting the number of valid pairs $(f(1), f(2))$ for each case. For each case, we multiplied the number of ways to choose $f(1), f(2),$ and $f(4)$ by the number of independent choices for $f(3)$ and $f(5)$ (which is $6 \times 6$). Finally, we summed the results from all mutually exclusive cases using the Addition Principle to obtain the total number of functions satisfying the given condition.

The final answer is $\boxed{360}$.