Let A = {0, 1, 2, 3, 4, 5, 6, 7}. Then the nu | vedclass

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(C) Given the condition $f(1) + f(2) = 3 - f(3)$,we can rewrite it as $f(1) + f(2) + f(3) = 3$.Since $f$ is a bijective function from $A$ to $A$,all v

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$720$

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(C) Given the condition $f(1) + f(2) = 3 - f(3)$,we can rewrite it as $f(1) + f(2) + f(3) = 3$.Since $f$ is a bijective function from $A$ to $A$,all values $f(1), f(2), f(3)$ must be distinct elements from the set $A = \{0, 1, 2, 3, 4, 5, 6, 7\}$.The only distinct non-negative integers that sum to $3$ are $0, 1,$ and $2$.Thus,the set of values $\{f(1), f(2), f(3)\}$ must be equal to the set $\{0, 1, 2\}$.The number of ways to assign these $3$ values to $f(1), f(2),$ and $f(3)$ is $3! = 6$.The remaining $5$ elements of the domain $\{0, 4, 5, 6, 7\}$ must be mapped to the remaining $5$ elements of the codomain $\{3, 4, 5, 6, 7\}$.The number of ways to map these is $5! = 120$.Therefore,the total number of bijective functions is $3! 	imes 5! = 6 	imes 120 = 720$.

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

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