Let  $R_1$ and  $R_2$ be relations on the set $\left\{1,2,\mathrm{.....},50\right\}$  such that $R_1=\left\{\left(p,p^n\right):pisaprimeandn\ge 0isan\mathrm{int}eger\right\}$
 and $R_2=\left\{\left(p,p^n\right):pisaprimeandn=0or1\right\}$.
Then, the number of elements in  $R_1-R_2$ is_________.
 

Given a set  $\left\{1,2,\mathrm{.....50}\right\}$
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 
$\left(2,2^0\right),\left(2,2^1\right)\mathrm{..}\left(2,2^5\right)$
$\left(3,3^0\right),\mathrm{......}\left(3,3^3\right)$
$\left(5,5^0\right),\mathrm{......}\left(5,5^2\right)$
$\left(7,7^0\right),\mathrm{.......}\left(7,7^2\right)$
$\left(11,{11}^0\right),\mathrm{........}\left(11,{11}^1\right)$
Every number of $P^n$  should lie in the given set 
.$\left\{1,2,\mathrm{.....50}\right\}$
Total number of relations are 
$n\left(R_1\right)=6+4+3+3+\left(2\times 10\right)=36$
Similarly for relation  $R_2,$
$\left(2,2^0\right),\left(2,2^1\right)\mathrm{......}\left(47,{47}^0\right),\left(47,{47}^1\right)$
Total number of relations are 
$n\left(R_2\right)=2\times 14=28$
Required difference 
$n\left(R_1\right)-n\left(R_2\right)=36-28=8$