Let A = {0, 1, 2, 3, 4, 5, 6, 7}. Then the number of bijective functions f : A → A such that f(1) +

Q: Wait, why do we reject the negative root or other combinations? Can't we pick other numbers?

Great question! Remember, our set $A$ consists only of non-negative integers $\{0, 1, 2, 3, 4, 5, 6, 7\}$. Since we need three distinct elements that sum to $3$, we are very limited. If we try to include any number greater than $2$, the sum will exceed $3$ (e.g., $0+1+3=4$). Thus, $\{0, 1, 2\}$ is the only possible set of images.

Q: Why must $f(1), f(2),$ and $f(3)$ be distinct?

Because the function $f$ is defined as a bijection from $A$ to $A$. A bijection is both injective (one-to-one) and surjective (onto). By definition, an injective function maps distinct inputs to distinct outputs. Since $1, 2,$ and $3$ are distinct inputs, their images $f(1), f(2),$ and $f(3)$ must be distinct.

Q: Why do we multiply the number of ways for the first three elements by the ways for the remaining five?

This is the Fundamental Principle of Counting. We have two independent stages: first, mapping the subset $\{1, 2, 3\}$ to $\{0, 1, 2\}$, and second, mapping the remaining five elements of the domain to the remaining five elements of the codomain. Since these choices are independent, we multiply the number of possibilities for each stage to get the total.

Question

Let $A = \{0, 1, 2, 3, 4, 5, 6, 7\}$. Then the number of bijective functions $f : A 	o A$ such that $f(1) + f(2) = 3 - f(3)$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given a set $A = \{0, 1, 2, 3, 4, 5, 6, 7\}$ and a bijective function $f: A 	o A$. A bijection requires that every element in the domain maps to a unique element in the codomain.

The constraint provided is $f(1) + f(2) = 3 - f(3)$. By rearranging this equation, we obtain the fundamental anchor for our mapping:

$$f(1) + f(2) + f(3) = 3$$

### The Search for the Subset

We must identify three distinct elements from set $A$ that sum to $3$. Since $f$ is a bijection, $f(1)$, $f(2)$, and $f(3)$ must be distinct values chosen from $A$.

If any of these values were greater than $2$, the sum would exceed $3$ because the smallest possible distinct values are $0$ and $1$. Therefore, the only possible set of images for the inputs $\{1, 2, 3\}$ is the set $\{0, 1, 2\}$.

Any other combination, such as $\{0, 1, 3\}$, results in a sum of $4$, which violates our constraint. Thus, the images of $1, 2,$ and $3$ must be a permutation of $\{0, 1, 2\}$.

### The Permutation of the First Three

We now determine the number of ways to map the inputs $\{1, 2, 3\}$ to the targets $\{0, 1, 2\}$. This is a classic permutation problem of $3$ distinct items into $3$ distinct slots.

The number of ways to arrange these is given by $3!$:

$$3! = 3 	imes 2 	imes 1 = 6$$

We have successfully accounted for all constrained mappings.

### The Remaining Dance

Next, we consider the remaining elements. We have used the inputs $\{1, 2, 3\}$ and the codomain values $\{0, 1, 2\}$.

The remaining elements in the domain are $\{0, 4, 5, 6, 7\}$, and the remaining elements in the codomain are $\{3, 4, 5, 6, 7\}$. Since $f$ must be a bijection, these $5$ elements must be mapped to each other in a one-to-one fashion.

The number of ways to map these $5$ elements is the number of permutations of $5$ items:

$$5! = 5 	imes 4 	imes 3 	imes 2 	imes 1 = 120$$

### The Final Synthesis

We have two independent sets of choices. According to the Fundamental Principle of Counting, we multiply the number of ways to map the first group by the number of ways to map the second group.

The total number of such bijective functions is:

$$6 	imes 120 = 720$$

The total number of bijective functions satisfying the given constraint is **720**.

Animated Solution for Mathematics - Functions: Let $A = {0, 1, 2, 3, 4, 5, 6, 7}$ . Then the number of bijective functions $f : A \to A$ such that $f (1) + f (2) = 3 - f (3)$ is equal to

Enter Numerical Value:

Visualized Solution

Understanding the Set $A$

Rearranging the Condition

Finding Possible Values

Permuting the First Three Elements

Identifying Remaining Elements

Permuting Remaining Elements

Calculating Total Functions

The Sigma Insight: Classification of Functions

Analyzing the Setup

The Search for the Subset

The Permutation of the First Three

The Remaining Dance

The Final Synthesis

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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={0,1,2,3,4,5,6,7}. Then the number of bijective functions f:A→A such that f(1)+f(2)=3−f(3) is equal to