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

There are 16 ordered pairs in $S \times S$. We write all these ordered pairs in 4 sets as follows. <br/><br/> $A=\{(1,1)\}$ <br/><br/> $B=\{(1,4),(2,4),(3,4)(4,4),(4,3),(4,2),(4,1)\}$ <br/><br/> $C=\{(1,3),(2,3),(3,3),(3,2),(3,1)\}$ <br/><br/> $D=\{(1,2),(2,2),(2,1)\}$ <br/><br/> All elements of set $B$ have image 4 and only element of $A$ has image 1. <br/><br/> All elements of set $C$ have image 3 or 4 and all elements of set $D$ have image 2 or 3 or 4 . <br/><br/> We will solve this question in two cases. <br/><br/> <b>Case I</b>: When no element of set $C$ has image 3. <br/><br/> Number of onto functions $=2$ (when elements of set $D$ have images 2 or 3$)$ <br/><br/> <b>Case II</b>: When atleast one element of set $C$ has image 3.<br/><br/> Number of onto functions $=\left(2^{3}-1\right)(1+2+2)$ <br/><br/> $=35$ <br/><br/> Total number of functions $=37$