Let S = {1, 2, 3, 4}. Then the number of elem | vedclass

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(A) Given $S = \{1, 2, 3, 4\}$. The condition $f(a, b) = f(b, a) \geq a$ implies:For $a=4$,$f(4, b) = f(b, 4) \geq 4$. Since the codomain is $S$,$f(4,

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$37$

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(A) Given $S = \{1, 2, 3, 4\}$. The condition $f(a, b) = f(b, a) \geq a$ implies:For $a=4$,$f(4, b) = f(b, 4) \geq 4$. Since the codomain is $S$,$f(4, b) = 4$ for all $b \in S$.For $a=3$,$f(3, b) = f(b, 3) \geq 3$. Thus $f(3, 3) \in \{3, 4\}$ and $f(3, 4) = f(4, 3) = 4$.For $a=2$,$f(2, b) = f(b, 2) \geq 2$. Thus $f(2, 2) \in \{2, 3, 4\}$,$f(2, 3) = f(3, 2) \in \{3, 4\}$,and $f(2, 4) = f(4, 2) = 4$.For $a=1$,$f(1, b) = f(b, 1) \geq 1$. Thus $f(1, 1) \in \{1, 2, 3, 4\}$,$f(1, 2) = f(2, 1) \in \{2, 3, 4\}$,$f(1, 3) = f(3, 1) \in \{3, 4\}$,and $f(1, 4) = f(4, 1) = 4$.To be onto,the range must be $\{1, 2, 3, 4\}$.Let $x_1 = f(1, 1)$,$x_2 = f(1, 2)$,$x_3 = f(1, 3)$,$x_4 = f(2, 2)$,$x_5 = f(2, 3)$,$x_6 = f(3, 3)$.$x_1 \in \{1, 2, 3, 4\}$,$x_2 \in \{2, 3, 4\}$,$x_3 \in \{3, 4\}$,$x_4 \in \{2, 3, 4\}$,$x_5 \in \{3, 4\}$,$x_6 \in \{3, 4\}$.Total functions satisfying the condition $= 4 	imes 3 	imes 2 	imes 3 	imes 2 	imes 2 = 288$.Using the Principle of Inclusion-Exclusion to ensure the function is onto:Let $P_i$ be the property that $i$ is not in the range.Total functions $= 288$.Functions missing $1$: $f(1, 1) \in \{2, 3, 4\} \implies 3 	imes 3 	imes 2 	imes 3 	imes 2 	imes 2 = 216$.Functions missing $2$: $f(1, 1) \in \{1, 3, 4\}, f(1, 2) \in \{3, 4\}, f(2, 2) \in \{3, 4\} \implies 3 	imes 2 	imes 2 	imes 2 	imes 2 	imes 2 = 96$.Functions missing $3$: $f(1, 1) \in \{1, 2, 4\}, f(1, 2) \in \{2, 4\}, f(1, 3) = 4, f(2, 2) \in \{2, 4\}, f(2, 3) = 4, f(3, 3) = 4 \implies 3 	imes 2 	imes 1 	imes 2 	imes 1 	imes 1 = 12$.After calculating intersections and applying $PIE$,the number of onto functions is $37$.

Let $S = \{1, 2, 3, 4\}$. Then the number of elements in the set $\{f: S \times S \rightarrow S : f \text{ is onto and } f(a, b) = f(b, a) \geq a; \forall (a, b) \in S \times S\}$ is

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