Let A = {1, 2, 3, 4, 5, 6}. The number of one | vedclass

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(NONE) Given $f: A 	o A$ is a one-one function where $A = \{1, 2, 3, 4, 5, 6\}$.We have the condition $f(2) + f(3) = 5$ and $f(3) \le 4$.The possible

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(NONE) Given $f: A 	o A$ is a one-one function where $A = \{1, 2, 3, 4, 5, 6\}$.We have the condition $f(2) + f(3) = 5$ and $f(3) \le 4$.The possible pairs $(f(2), f(3))$ are $(1, 4), (4, 1), (2, 3), (3, 2)$.Since $f$ is one-one,$f(1), f(2), f(3)$ must be distinct.Case $1$: $(f(2), f(3)) = (1, 4)$. Then $f(1) \in A \setminus \{1, 4\} = \{2, 3, 5, 6\}$. Given $f(1) \ge 3$,so $f(1) \in \{3, 5, 6\}$ ($3$ choices). Remaining $3$ elements can be mapped in $3! = 6$ ways. Total $= 3 	imes 6 = 18$.Case $2$: $(f(2), f(3)) = (4, 1)$. Then $f(1) \in A \setminus \{4, 1\} = \{2, 3, 5, 6\}$. Given $f(1) \ge 3$,so $f(1) \in \{3, 5, 6\}$ ($3$ choices). Remaining $3$ elements can be mapped in $3! = 6$ ways. Total $= 3 	imes 6 = 18$.Case $3$: $(f(2), f(3)) = (2, 3)$. Then $f(1) \in A \setminus \{2, 3\} = \{1, 4, 5, 6\}$. Given $f(1) \ge 3$,so $f(1) \in \{4, 5, 6\}$ ($3$ choices). Remaining $3$ elements can be mapped in $3! = 6$ ways. Total $= 3 	imes 6 = 18$.Case $4$: $(f(2), f(3)) = (3, 2)$. Then $f(1) \in A \setminus \{3, 2\} = \{1, 4, 5, 6\}$. Given $f(1) \ge 3$,so $f(1) \in \{4, 5, 6\}$ ($3$ choices). Remaining $3$ elements can be mapped in $3! = 6$ ways. Total $= 3 	imes 6 = 18$.Total number of functions $= 18 + 18 + 18 + 18 = 72$.

Let $A = \{1, 2, 3, 4, 5, 6\}$. The number of one-one functions $f: A \to A$ such that $f(1) \ge 3, f(3) \le 4$ and $f(2) + f(3) = 5$,is ————

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