Key Concepts and Formulas

• Bijective Function: A function $f: A \to A$ is bijective if it is both injective (one-to-one) and surjective (onto). For a finite set $A$, a function is bijective if and only if it is a permutation of the elements of $A$.
• Number of Bijective Functions: The number of bijective functions from a set of $n$ elements to itself is $n!$.
• Combinations and Permutations: When selecting and arranging distinct items, permutations are used. The number of permutations of $k$ items chosen from $n$ is $P(n, k) = \frac{n!}{(n-k)!}$. When the order does not matter, combinations are used, $C(n, k) = \frac{n!}{k!(n-k)!}$.

Step-by-Step Solution

Step 1: Understand the Problem and the Condition

We are given a set $A = \{0, 1, 2, 3, 4, 5, 6, 7\}$, which has $n=8$ elements. We need to find the number of bijective functions $f: A \to A$ such that $f(1) + f(2) = 3 - f(3)$. First, let's rearrange the given condition to make it more manageable: $f(1) + f(2) = 3 - f(3)$ $f(1) + f(2) + f(3) = 3$ Since $f$ is a bijective function, the values $f(1)$, $f(2)$, and $f(3)$ must be distinct elements from the set $A$.

Step 2: Determine the Possible Values for $f(1)$, $f(2)$, and $f(3)$

We need to find three distinct elements from the set $A = \{0, 1, 2, 3, 4, 5, 6, 7\}$ whose sum is 3. Let these three distinct elements be $x, y, z \in A$. We are looking for solutions to $x + y + z = 3$, where $x \neq y$, $y \neq z$, and $x \neq z$. Since all elements in $A$ are non-negative integers, we can systematically check possible combinations. The smallest possible sum of three distinct non-negative integers is $0 + 1 + 2 = 3$. Let's consider if any other combination is possible. If we try to include any element greater than 2, for instance, if one of the elements is 3, the sum would be at least $0 + 1 + 3 = 4$, which is greater than 3. Therefore, the only possible set of three distinct elements from $A$ that sums to 3 is $\{0, 1, 2\}$. Thus, the set $\{f(1), f(2), f(3)\}$ must be equal to $\{0, 1, 2\}$.

Step 3: Count the Number of Ways to Assign Values to $f(1)$, $f(2)$, and $f(3)$

We know that the values $f(1)$, $f(2)$, and $f(3)$ must be a permutation of the set $\{0, 1, 2\}$. We need to assign these three values to the three distinct inputs $1, 2, 3$.

• $f(1)$ can take any of the 3 values (0, 1, or 2).
• Once $f(1)$ is chosen, $f(2)$ can take any of the remaining 2 values.
• Finally, $f(3)$ must take the last remaining value. The number of ways to assign these values is the number of permutations of 3 elements, which is $3!$. $3! = 3 \times 2 \times 1 = 6$ There are 6 ways to assign the values $\{0, 1, 2\}$ to $f(1), f(2), f(3)$.

Step 4: Count the Number of Ways to Map the Remaining Elements

After assigning values to $f(1)$, $f(2)$, and $f(3)$, we have used 3 elements from the domain ($1, 2, 3$) and 3 elements from the codomain ($\{0, 1, 2\}$). The remaining elements in the domain are $A \setminus \{1, 2, 3\} = \{0, 4, 5, 6, 7\}$. There are $8 - 3 = 5$ remaining elements in the domain. The remaining elements in the codomain are $A \setminus \{0, 1, 2\} = \{3, 4, 5, 6, 7\}$. There are $8 - 3 = 5$ remaining elements in the codomain. Since $f$ must be a bijective function, the remaining 5 elements of the domain must be mapped bijectively to the remaining 5 elements of the codomain. The number of ways to do this is the number of permutations of 5 elements, which is $5!$. $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

Step 5: Calculate the Total Number of Bijective Functions

To find the total number of bijective functions satisfying the given condition, we multiply the number of ways to map the first three elements by the number of ways to map the remaining elements. Total number of functions = (Number of ways to map $\{1, 2, 3\}$) $\times$ (Number of ways to map $\{0, 4, 5, 6, 7\}$) Total number of functions = $3! \times 5!$ Total number of functions = $6 \times 120 = 720$

Common Mistakes & Tips

• Forgetting Distinctness: Always remember that for a bijective function, $f(x_1) \neq f(x_2)$ if $x_1 \neq x_2$. This is crucial for identifying the possible values that sum to 3.
• Overlooking the Codomain: Ensure that the values assigned to $f(1), f(2), f(3)$ are indeed elements of the set $A$.
• Treating it as a Simple Sum: The problem is not just about finding sets of numbers that sum to 3. It's about assigning specific domain elements to specific codomain elements while maintaining bijectivity.

Summary

The problem requires us to find the number of bijective functions $f: A \to A$ satisfying $f(1) + f(2) + f(3) = 3$. By leveraging the property that $f(1), f(2), f(3)$ must be distinct elements of $A$, we deduce that the only possible set of values for $\{f(1), f(2), f(3)\}$ is $\{0, 1, 2\}$. There are $3!$ ways to assign these values to $f(1), f(2), f(3)$. The remaining $5$ elements of the domain must be mapped bijectively to the remaining $5$ elements of the codomain, which can be done in $5!$ ways. The total number of such functions is the product of these two quantities, $3! \times 5! = 720$.

The final answer is $\boxed{720}$.