Let $R_1$ and $R_2$ be relations on the set $\{1,2,\mathrm{..........},50\}$ such that $R_1=\{(p,p^n):p$ is a prime and $n\ge 0$ is an integer} and $R_2=\{(p,p^n):p\mathrm{is}\mathrm{a}\mathrm{prime}\mathrm{and}n=0\mathrm{or}1\}.$ Then $3$times the number of elements in $R_1-R_2$ is__________. 

Given a set $\{1,2,\mathrm{......50}\}$

Possible choices of P are $2,3,5,7,11,13,17,23,29,31,37,41,43$ and 47.

So, we can calculate no. of elements in $R_1$ as $(2,2^0),(2,2^1)\mathrm{......}(2,2^5)$

$(3,3^0),\mathrm{........}(3,3^3)$

$(5,5^0),\mathrm{.......}(5,5^2)$

$(7,7^0),\mathrm{.....}(7,7^2)$

$(11,{11}^0),\mathrm{......}(11,{11}^1)$

Every number of $P^n$ should lie in the given set $\{1,2,\mathrm{.....50}\}.$

Total number of relations are  $n\left(R_1\right)=6+4+3+(2\times 10)=36$

Similarly for relation $R_2, (2,2^0),(2,2^1)\mathrm{......}(47,{47}^0),(47,{47}^1)$

Total number of relations are $n\left(R_2\right)=2\times 14=28$

Required difference $n\left(R_1\right)-n\left(R_2\right)=36-28=8$