We need to count all functions $f:S\times S\to S,\qquad S=\{1,2,3,4\}$ such that:

1. $f$ is symmetric:
   $f(a,b)=f(b,a)$
2. $f(a,b)\ge a$ for all $(a,b)\in S\times S$
3. $f$ is onto $S$.

1. Use symmetry and the inequality

Since $f(a,b)=f(b,a),$ and also the condition says for every ordered pair $(a,b)$, $f(a,b)\ge a.$ Applying this to $(b,a)$ gives $f(b,a)\ge b.$ But $f(a,b)=f(b,a)$, so actually $f(a,b)\ge \max(a,b).$

So the value at $(a,b)$ must be at least $\max(a,b)$.

Because of symmetry, it is enough to choose values on pairs with $a\le b$. These are: $(1,1),(1,2),(1,3),(1,4),(2,2),(2,3),(2,4),(3,3),(3,4),(4,4).$ So there are $10$ independent positions.

2. Allowed values at each independent pair

For each pair $(a,b)$ with $a\le b$, we need $f(a,b)\in \{b,b+1,\dots,4\}$ because $\max(a,b)=b$.

Thus:

• $(1,1)$: values in $\{1,2,3,4\}$ → 4 choices
• $(1,2)$: values in $\{2,3,4\}$ → 3 choices
• $(1,3)$: values in $\{3,4\}$ → 2 choices
• $(1,4)$: values in $\{4\}$ → 1 choice
• $(2,2)$: values in $\{2,3,4\}$ → 3 choices
• $(2,3)$: values in $\{3,4\}$ → 2 choices
• $(2,4)$: values in $\{4\}$ → 1 choice
• $(3,3)$: values in $\{3,4\}$ → 2 choices
• $(3,4)$: values in $\{4\}$ → 1 choice
• $(4,4)$: values in $\{4\}$ → 1 choice

Hence total number of symmetric functions satisfying the inequality is $4\cdot 3\cdot 2\cdot 1\cdot 3\cdot 2\cdot 1\cdot 2\cdot 1\cdot 1=144.$

Now we impose the onto condition.

3. Onto condition

Since codomain is $S=\{1,2,3,4\}$, onto means all four values $1,2,3,4$ must occur.

But note:

• Value $4$ always occurs, because $f(4,4)\ge 4 \implies f(4,4)=4.$ So we only need to ensure that $1,2,3$ also appear somewhere.

4. When does value 1 appear?

A value $1$ can occur only if $\max(a,b)\le 1$, i.e. only at $(1,1)$. So for onto, $f(1,1)=1.$ This is necessary.

So fix $f(1,1)=1.$ Then the remaining independent entries are: $(1,2),(1,3),(1,4),(2,2),(2,3),(2,4),(3,3),(3,4),(4,4).$ Their choices give $3\cdot 2\cdot 1\cdot 3\cdot 2\cdot 1\cdot 2\cdot 1\cdot 1=36$ functions.

Among these, we must ensure value $2$ and value $3$ both occur.

5. Count functions with $f(1,1)=1$ and both 2 and 3 appearing

Let:

• $A$ = set of functions (among these 36) in which value $2$ does not appear.
• $B$ = set of functions in which value $3$ does not appear.

Required count: $36-|A|-|B|+|A\cap B|.$

Count $|A|$ : no value 2 anywhere

Positions where 2 could appear are exactly those with allowed set containing 2: $(1,2), (2,2).$ If 2 is forbidden:

• $(1,2)$ must be $3$ or $4$ → 2 choices
• $(2,2)$ must be $3$ or $4$ → 2 choices Other positions unchanged:
• $(1,3)$: 2 choices
• $(1,4)$: 1 choice
• $(2,3)$: 2 choices
• $(2,4)$: 1 choice
• $(3,3)$: 2 choices
• $(3,4)$: 1 choice
• $(4,4)$: 1 choice

Thus $|A|=2\cdot 2\cdot 2\cdot 1\cdot 2\cdot 1\cdot 2\cdot 1\cdot 1=32.$

Count $|B|$ : no value 3 anywhere

Positions where 3 could appear are: $(1,2),(1,3),(2,2),(2,3),(3,3).$ If 3 is forbidden:

• $(1,2)$: only $2$ or $4$ → 2 choices
• $(1,3)$: only $4$ → 1 choice
• $(2,2)$: only $2$ or $4$ → 2 choices
• $(2,3)$: only $4$ → 1 choice
• $(3,3)$: only $4$ → 1 choice Other fixed/unchanged:
• $(1,4),(2,4),(3,4),(4,4)$ all have 1 choice each.

Thus $|B|=2\cdot 1\cdot 1\cdot 2\cdot 1\cdot 1\cdot 1\cdot 1\cdot 1=4.$

Count $|A\cap B|$ : neither 2 nor 3 appears

Then only values allowed are 1 and 4, but after $(1,1)=1$, all other positions must be 4. Check feasibility:

• $(1,2)$ cannot be 1, so must be 4
• $(2,2)$ must be 4
• all others also 4 Hence exactly 1 function. So $|A\cap B|=1.$

Therefore required number is $36-32-4+1=1.$

6. Direct check of the unique onto function

Indeed, if onto is required:

• to get value 1, must have $f(1,1)=1$;
• to get value 2, since only $(1,2)$ or $(2,2)$ can give 2, at least one of these must be 2;
• to get value 3, note every place capable of giving 3 except $(3,3)$ may be used, but if either $(1,2)$ or $(2,2)$ is 2, and onto must still happen, let us inspect carefully.

A simpler observation is:

• value 3 can occur at $(1,3),(2,3),(3,3),(1,2),(2,2)$. So multiple onto functions may seem possible, but inclusion-exclusion above counts exactly 1. Let us verify by explicit reasoning:

If value 2 appears, it must be at $(1,2)$ or $(2,2)$. If value 3 appears, it can be at one of the 5 positions above. Thus clearly more than 1 onto function should exist. So let us re-check the inclusion-exclusion count.

7. Re-check counting carefully

After fixing $f(1,1)=1$, the 6 nontrivial variable positions are actually:

• $(1,2)$: 3 choices
• $(1,3)$: 2 choices
• $(2,2)$: 3 choices
• $(2,3)$: 2 choices
• $(3,3)$: 2 choices and the fixed ones are
• $(1,4)=(2,4)=(3,4)=(4,4)=4$. So total is still $3\cdot 2\cdot 3\cdot 2\cdot 2=72,$ not 36. The previous total 36 mistakenly omitted one factor 2.

Now redo inclusion-exclusion.

Total with $f(1,1)=1$

$N=3\cdot 2\cdot 3\cdot 2\cdot 2=72.$

No 2 anywhere

Then:

• $(1,2)$: $\{3,4\}$ → 2 choices
• $(2,2)$: $\{3,4\}$ → 2 choices
• $(1,3)$: 2 choices
• $(2,3)$: 2 choices
• $(3,3)$: 2 choices Hence $|A|=2\cdot 2\cdot 2\cdot 2\cdot 2=32.$

No 3 anywhere

Then:

• $(1,2)$: $\{2,4\}$ → 2 choices
• $(1,3)$: $\{4\}$ → 1 choice
• $(2,2)$: $\{2,4\}$ → 2 choices
• $(2,3)$: $\{4\}$ → 1 choice
• $(3,3)$: $\{4\}$ → 1 choice Hence $|B|=2\cdot 1\cdot 2\cdot 1\cdot 1=4.$

Neither 2 nor 3 anywhere

Then:

• $(1,2)=4,, (1,3)=4,, (2,2)=4,, (2,3)=4,, (3,3)=4,