Let $\mathrm{S}=\{1,2,3,4\}$. Then the number of element in the set $\{\mathrm{f}:\mathrm{S}\times \mathrm{S}\to \mathrm{S}:\mathrm{f}\text{ is onto and }\mathrm{f}(\mathrm{a},\mathrm{b})=\mathrm{f}(\mathrm{b},\mathrm{a})⩾\mathrm{a}\mathrm{\forall }(\mathrm{a},\mathrm{b})\in \mathrm{S}\times \mathrm{S}\}$ is _____________.

There are 16 order pairs in S X S 
$$\begin{gathered} A=\left\{(1,1)\right\} \\ B=\left\{\left(1,4\right), \left(2,4\right),\left(3,4\right),(4,4),(4,3),(4,2),(4,1)\right\} \\ C=\left\{\left(1,3\right),(2,3),(3,3),(3,2),(3,1)\right\} \\ D=\left\{(1,2),(2,2),(2,1)\right\} \\ All the elements of set B have image 4 and only element of A has image 1 \end{gathered}$$
$$\begin{gathered} \mathrm{All} \mathrm{the} \mathrm{elements} \mathrm{of} \mathrm{setC} \mathrm{have} \mathrm{image} 3 \mathrm{or} 4 \mathrm{and} \mathrm{all} \mathrm{elements} \mathrm{of} \mathrm{set} \mathrm{D} \mathrm{have} \\ \mathrm{image} 2 \mathrm{or} 3 \mathrm{or} 4 \\ \mathrm{Case}-1: \mathrm{No} \mathrm{element} \mathrm{in} \mathrm{set} \mathrm{C} \mathrm{has} \mathrm{image}3 \\ \mathrm{Number} \mathrm{of} \mathrm{onto} \mathrm{functions} = 2(\mathrm{when} \mathrm{elements} \mathrm{of} \mathrm{setD} \mathrm{have} \mathrm{images} 2 \mathrm{or} 3) \\ \mathrm{Case}-2: \mathrm{when} \mathrm{atleast} \mathrm{one} \mathrm{element} \mathrm{of} \mathrm{set} \mathrm{c} \mathrm{has} \mathrm{image} 3 \\ \mathrm{Number} \mathrm{of} \mathrm{onto} \mathrm{functions} =(2^3-1)(1+2+2)=35 \\ \mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{functions}= 37 \end{gathered}$$