Key Concepts and Formulas

• Functions: A function $f: A \rightarrow B$ assigns exactly one element from the codomain $B$ to each element in the domain $A$.
• Counting Principles: The multiplication principle (for independent choices) and the addition principle (for mutually exclusive cases) are fundamental.
• Systematic Enumeration: For constraints involving sums or products, breaking down the problem into cases based on the possible values of one or more variables is a common strategy.

Step-by-Step Solution

Step 1: Understanding the Problem and Identifying Independent Choices

We are asked to find the number of functions $f:\{1,2,3,4\} \rightarrow\{1,2,3,4,5,6\}$ such that $f(1)+f(2)=f(3)$. The domain is $\{1,2,3,4\}$ and the codomain is $\{1,2,3,4,5,6\}$. This means $f(1), f(2), f(3), f(4)$ must all be integers between 1 and 6, inclusive.

The constraint $f(1)+f(2)=f(3)$ involves the function values at 1, 2, and 3. The function value $f(4)$ is not mentioned in any constraint. Therefore, the choice of $f(4)$ is independent of the choices for $f(1), f(2),$ and $f(3)$. Since $f(4)$ can be any of the 6 values in the codomain, there are 6 possible choices for $f(4)$. We can calculate the number of ways to satisfy $f(1)+f(2)=f(3)$ for $f(1), f(2), f(3)$ in the codomain, and then multiply this count by 6 to account for all possibilities of $f(4)$.

Step 2: Analyzing the Constraint $f(1)+f(2)=f(3)$

We need to find the number of ordered triples $(f(1), f(2), f(3))$ such that $f(1), f(2), f(3) \in \{1,2,3,4,5,6\}$ and $f(1)+f(2)=f(3)$.

Let's consider the possible values for $f(3)$.

• The minimum value of $f(1)+f(2)$ is $1+1=2$. So, $f(3)$ must be at least 2.
• The maximum value of $f(1)+f(2)$ is $6+6=12$. Since $f(3)$ must also be in the codomain $\{1,2,3,4,5,6\}$, the possible values for $f(3)$ are $\{2,3,4,5,6\}$. We will now enumerate the possibilities for $(f(1), f(2))$ for each of these values of $f(3)$.

Step 3: Enumerating Pairs $(f(1), f(2))$ for each possible value of $f(3)$

We will consider each possible value of $f(3)$ from the set $\{2,3,4,5,6\}$ and find the number of pairs $(f(1), f(2))$ from $\{1,2,3,4,5,6\} \times \{1,2,3,4,5,6\}$ that satisfy $f(1)+f(2)=f(3)$.

• Case 1: $f(3)=2$ We need $f(1)+f(2)=2$. The only possible pair is $(f(1), f(2)) = (1,1)$. Number of pairs: 1.

• Case 2: $f(3)=3$ We need $f(1)+f(2)=3$. The possible pairs are $(1,2)$ and $(2,1)$. Number of pairs: 2.

• Case 3: $f(3)=4$ We need $f(1)+f(2)=4$. The possible pairs are $(1,3), (2,2), (3,1)$. Number of pairs: 3.

• Case 4: $f(3)=5$ We need $f(1)+f(2)=5$. The possible pairs are $(1,4), (2,3), (3,2), (4,1)$. Number of pairs: 4.

• Case 5: $f(3)=6$ We need $f(1)+f(2)=6$. The possible pairs are $(1,5), (2,4), (3,3), (4,2), (5,1)$. Number of pairs: 5.

The total number of valid triples $(f(1), f(2), f(3))$ satisfying the condition is the sum of the counts from these cases: $1 + 2 + 3 + 4 + 5 = 15$.

Step 4: Calculating the Total Number of Functions

For each of the 15 valid triples $(f(1), f(2), f(3))$, there are 6 independent choices for $f(4)$. Therefore, the total number of functions is the number of valid triples multiplied by the number of choices for $f(4)$.

Total number of functions = (Number of valid $(f(1), f(2), f(3))$ triples) $\times$ (Number of choices for $f(4)$) $= 15 \times 6$ $= 90$

Common Mistakes & Tips

• Forgetting $f(4)$: Always check if all elements in the domain have their function values accounted for. If some are unconstrained, their possibilities multiply the total count.
• Treating $(f(1), f(2))$ as unordered: The order matters; $(1,2)$ is a different assignment for $(f(1), f(2))$ than $(2,1)$.
• Ignoring Codomain Restrictions: Ensure that all function values, especially those resulting from sums, remain within the specified codomain.

Summary

We determined the number of functions by first analyzing the constraint $f(1)+f(2)=f(3)$. We found that $f(3)$ could take values from 2 to 6. By systematically enumerating the pairs $(f(1), f(2))$ for each possible value of $f(3)$, we found there were 15 ways to satisfy the constraint for $f(1), f(2), f(3)$. Since $f(4)$ could be any of the 6 values in the codomain independently, we multiplied the count of valid $(f(1), f(2), f(3))$ assignments by 6. This resulted in a total of $15 \times 6 = 90$ functions.

The final answer is $\boxed{90}$, which corresponds to option (B).