Let R₁ and R₂ be relations on the set {1, 2, …….., 50} such that

${\mathrm{R}}_1=\left\{\left(\mathrm{p},{\mathrm{p}}^{\mathrm{n}}\right):\mathrm{p}\right.$is a prime and $\mathrm{n}\ge 0$ is an integer} and

${\mathrm{R}}_2=\left\{\left(\mathrm{p},{\mathrm{p}}^{\mathrm{n}}\right):\mathrm{p}\text{ is a prime and }\mathrm{n}=0\text{ or }1\right\}$

Then, the number of elements in R₁ - R₂ is _______________.

$\text{ Here, }\mathrm{p},{\mathrm{p}}^{\mathrm{n}}\in \{1,2,\dots 50\}$$\text{ Now p can take values }$

$2,3,5,7,11,13,17,23,29,31,37,41,43\text{ and }47.$

Elements in R ₁are 

$\begin{array}{r}\left(2,2^{\circ }\right),\left(2,2^1\right)\dots \left(2,2^5\right)\\ \left(3,3^{\circ }\right),\dots \left(3,3^3\right)\\ \left(5,5^{\circ }\right),\dots \left(5,5^2\right)\\ \left(7,7^{\circ }\right),\dots \left(7,7^2\right)\\ \left(11,{11}^{\circ }\right),\dots \left(11,{11}^1\right)\end{array}$

$\therefore \mathrm{n}\left({\mathrm{R}}_1\right)=6+4+3+3+(2\times 10)=36$

Similarly for R2$\begin{array}{r}\left(2,2^{\circ }\right),\left(2,2^1\right)\left(47,{47}^{\circ }\right),\left(47,{47}^1\right)\\ \end{array}$

$\begin{array}{l}\therefore \mathrm{n}\left({\mathrm{R}}_2\right)=2\times 14=28\\ \therefore \mathrm{n}\left({\mathrm{R}}_1\right)-\mathrm{n}\left({\mathrm{R}}_2\right)=36-28=8\end{array}$