Let R be a relation from the set $\{1,2,3,\dots ,60\}$ to itself such that $\mathrm{R}=\left\{\right(\mathrm{a},\mathrm{b}):\mathrm{b}=\mathrm{pq}$,where $\mathrm{p},\mathrm{q}\ge 3$are prime numbers}. Then, the number of elements in R is :

$\mathrm{R}=\left\{\right(\mathrm{a},\mathrm{b}):\mathrm{b}=\mathrm{pq}$ where $\mathrm{p},\mathrm{q}\ge 3$ are prime numbers} 
Number of possible values of a =.60
For $\mathrm{b}=\mathrm{pq},\text{ if }\mathrm{p}=3,\mathrm{q}=3,5,7,11,13,17,19$
If $\mathrm{p}=5,\mathrm{q}=5,7,11;\text{ if }\mathrm{p}=7,\mathrm{q}=7$
$\therefore$Number of elements in $\mathrm{R}=60(7+3+1)=660$