Let A = {1, 2, 3, 4, 5} and B = {1, 2, 3, 4, 5, 6}. Then the number of functions f : A → B satisfyin

Q: Wait, why can't we just pick values for $f(1), f(2), f(4)$ independently? Why is there a constraint?

In a general function $f: A \to B$, each element in the domain can map to any element in the codomain. However, the problem imposes a specific algebraic constraint: $f(1) + f(2) = f(4) - 1$. This turns a 'free choice' problem into a 'constrained' problem. We are essentially looking for the subset of all possible functions that satisfy this specific geometric relationship.

Q: Why do we stop checking cases at $f(1) = 4$? Can't $f(1)$ be $5$?

That is a fantastic question! Let's look at the math. If $f(1) = 5$, then the equation becomes $5 + f(2) = f(4) - 1$, which simplifies to $f(4) = f(2) + 6$. Since the minimum value for $f(2)$ is $1$, this would force $f(4) \ge 7$. But our codomain $B$ only goes up to $6$. Thus, $f(4)$ would fall outside the codomain, making such a function impossible. We stop at $4$ because that is the physical limit of our universe (the codomain).

Q: What about $f(3)$ and $f(5)$? Do they affect the constraint?

They are the 'free agents' of this problem! Because the constraint only involves $f(1), f(2),$ and $f(4)$, the elements $3$ and $5$ in the domain are completely unburdened. They can map to any of the $6$ elements in $B$. This is why we multiply our final count by $6 \times 6$ at the end. Always look for the variables that are 'free' versus those that are 'bound'.

Question

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given a domain $A = \{1, 2, 3, 4, 5\}$ and a codomain $B = \{1, 2, 3, 4, 5, 6\}$. We need to determine the number of functions $f: A 	o B$ that satisfy the specific constraint:

$$f(1) + f(2) = f(4) - 1$$

### Decoding the Constraint

The equation $f(1) + f(2) = f(4) - 1$ acts as a rule binding the outputs of the function. We can rewrite this as:

$$f(4) = f(1) + f(2) + 1$$

This implies that $f(4)$ is entirely determined by the values chosen for $f(1)$ and $f(2)$. Once $f(1)$ and $f(2)$ are fixed, $f(4)$ is automatically constrained.

### The Boundary Trap

We must ensure that $f(4)$ remains within the codomain $B$. Since the maximum value in $B$ is $6$, we must satisfy the condition $f(4) \le 6$.

Substituting our constraint into this inequality, we get:

$$f(1) + f(2) + 1 \le 6$$

This simplifies to the fundamental constraint:

$$f(1) + f(2) \le 5$$

This inequality defines the "universe" of valid possibilities for the pair $(f(1), f(2))$. If the sum exceeds $5$, $f(4)$ would exceed $6$, which is impossible.

### Systematic Enumeration

We now count the valid pairs $(f(1), f(2))$ that satisfy $f(1) + f(2) \le 5$, where $f(1), f(2) \in \{1, 2, 3, 4, 5, 6\}$.

*   If $f(1) = 1$, then $f(2) \le 4$. This yields $4$ valid pairs: $(1,1), (1,2), (1,3), (1,4)$.
*   If $f(1) = 2$, then $f(2) \le 3$. This yields $3$ valid pairs: $(2,1), (2,2), (2,3)$.
*   If $f(1) = 3$, then $f(2) \le 2$. This yields $2$ valid pairs: $(3,1), (3,2)$.
*   If $f(1) = 4$, then $f(2) \le 1$. This yields $1$ valid pair: $(4,1)$.

Summing these possibilities, the total number of valid triplets $(f(1), f(2), f(4))$ is:

$$4 + 3 + 2 + 1 = 10$$

### The Freedom of the Unconstrained

The elements $3$ and $5$ from the domain $A$ are not mentioned in the constraint. Therefore, $f(3)$ and $f(5)$ are completely free to map to any of the $6$ elements in $B$.

Each of these two elements has $6$ choices. By the **Fundamental Principle of Counting**, we multiply the number of ways to map the constrained elements by the number of ways to map the free elements:

$$	ext{Total Functions} = 10 	imes 6 	imes 6 = 360$$

The total number of such functions is **360**.

Animated Solution for Mathematics - Functions: 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

Enter Numerical Value:

Visualized Solution

Visualizing Sets $A$ and $B$

Analyzing the Constraint

Establishing the Bounds

Case 1: $f (1) = 1$

Case 2: $f (1) = 2$

Cases 3 & 4: $f (1) = 3$ and $f (1) = 4$

Total Pairs for $(f (1), f (2))$

Handling Remaining Elements

Final Calculation

The Sigma Insight: Classification of Functions

Analyzing the Setup

Decoding the Constraint

The Boundary Trap

Systematic Enumeration

The Freedom of the Unconstrained

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Animated Solution for Mathematics - Functions: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A={1,2,3,4,5} and B={1,2,3,4,5,6}. Then the number of functions f:A→B satisfying f(1)+f(2)=f(4)−1 is equal to

Enter Numerical Value:

Visualized Solution

Visualizing Sets .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Analyzing the Constraint

Establishing the Bounds

Case 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(1)=1

Case 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(1)=2

Cases 3 & 4: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(1)=3 and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(1)=4

Total Pairs for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (f(1),f(2))