Animated Solution for Mathematics - Functions: Let $S=\{1,2,3,4\}$. Then the number of elements in the set $\{f:S\times S\to S:f\text{ is onto and }f(a,b)=f(b,a)\ge a\forall (a,b)\in S\times S\}$ is ______.

We are given a set $S=\{1,2,3,4\}$ and a function $f:S\times S\to S$. The function must satisfy two primary conditions: symmetry, defined as $f(a,b)=f(b,a)$, and a lower bound constraint, $f(a,b)\ge \max (a,b)$.

Because the function is symmetric, we only need to determine the values for the $10$ pairs where $a\le b$. The domain is effectively reduced to the set of pairs $\{(1,1),(1,2),(1,3),(1,4),(2,2),(2,3),(2,4),(3,3),(3,4),(4,4)\}$.

The condition $f(a,b)\ge \max (a,b)$ acts as a strict constraint on the possible outputs for each pair. For pairs where $\max (a,b)=4$, specifically $(1,4),(2,4),(3,4),$ and $(4,4)$, the output must be $\ge 4$.

Since the codomain is limited to $\{1,2,3,4\}$, the only possible output for these four pairs is $4$. Thus, these values are fixed:

For the function to be surjective (onto), every element in $S$ must appear as an output. The value $1$ can only be produced by $f(1,1)$ because for any other pair $(a,b)$, $\max (a,b)>1$, forcing $f(a,b)\ge \max (a,b)>1$.

Therefore, we must have $f(1,1)=1$. This leaves us with two remaining groups of pairs to assign: Group $Y=\{f(1,2),f(2,2)\}$ where the max is $2$, allowing outputs in $\{2,3,4\}$. Group $X=\{f(1,3),f(2,3),f(3,3)\}$ where the max is $3$, allowing outputs in $\{3,4\}$.

The total number of ways to assign these values without considering the surjectivity of $2$ and $3$ is:

We must ensure that $2$ and $3$ are included in the range. Let $A$ be the set of functions where $2$ is missing, and $B$ be the set of functions where $3$ is missing. We seek the total minus $|A\cup B|=|A|+|B|-|A\cap B|$.

If $2$ is missing (Set $A$), Group $Y$ is restricted to $\{3,4\}$ ($2^2=4$ ways) and Group $X$ is restricted to $\{3,4\}$ ($2^3=8$ ways). Thus, $|A|=4\times 8=32$.

If $3$ is missing (Set $B$), Group $Y$ is restricted to $\{2,4\}$ ($2^2=4$ ways) and Group $X$ is restricted to $\{4\}$ ($1^3=1$ way). Thus, $|B|=4\times 1=4$.

If both are missing ($A\cap B$), Group $Y$ is restricted to $\{4\}$ ($1^2=1$ way) and Group $X$ is restricted to $\{4\}$ ($1^3=1$ way). Thus, $|A\cap B|=1\times 1=1$.

The number of invalid functions is:

Subtracting the invalid functions from the total combinations, we find the number of surjective functions:

The total number of such functions is 37.