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

Now  numbers  of   possible  values   of  $a =60, for b=pq$
If    $p=3, q=3,5,7,11,13,17,19$
Total  cases =7 
If  p=  5q =5,7,11                  total case =3 
If p=7q  =7                            total case =1
Possible  value of    7+3+1=11 and a  =60
Total cases  $a\times b=60\times 11=660$