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

R₁ = {(p, pⁿ) : p is a prime and n $\ge$ 0 is an integer} and

R₂ = {(p, pⁿ) : p is a prime and n = 0 or 1}.

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

Given, ${R}_1=\left\{\left(p, p^n\right): p\right.$ is a Prime and $n \geq 0$ is an integer $\}$ <br/><br/>and, set $A=\{1,2,3 \ldots \ldots .50\}$ <br/><br/>$p$ is a Prime number which can take 15 values $2,3,5,7,11,13,17,19,23,29,31,37,41,43$ and 47 <br/><br/>$\therefore$ We can calculate no. of elements in $\mathrm{R}_1$ <br/><br/>$$ \mathrm{R}_1=\left(2,2^0\right),\left(2,2^1\right),\left(2,2^2\right)\left(2,2^3\right) \ldots \ldots\left(2,2^5\right)=6$$ number of ordered pairs <br/><br/>$\left(3,3^0\right),\left(3,3^1\right),\left(3,3^2\right) \ldots \ldots . .\left(3,3^3\right)=4$ number of order paris <br/><br/>$\left(5,5^0\right),\left(5,5^1\right),\left(5,5^2\right) \ldots \ldots \ldots . .=3$ number of order paris <br/><br/>$\left(7,7^0\right) \ldots \ldots .\left(7,7^2\right) \ldots \ldots \ldots=3$ number of order paris <br/><br/>$\left(11,11^0\right)$ and $\left(11,11^1\right)=2$ number of order paris <br/><br/>$\left(13,13^0\right)$ and $\left(13,13^1\right)=2$ number of order paris <br/><br/>$\therefore$ For the 11 prime numbers ($11,13,17,19,23,29,31,37,41,43$ and 47), $n$ can only be 0, 1 (two pairs each). <br/><br/>$\therefore n\left(\mathrm{R}_1\right)=6+4+3+3+(2 \times 11)=38$ <br/><br/>$\mathrm{R}_2=\left(p, p^n\right) $, where n = 0 or 1 <br/><br/>$$ \left(2,2^0\right),\left(2,2^1\right)\left(3,3^0\right)\left(3,3^1\right) \ldots . .\left(47,47^0\right)\left(47,47^1\right) $$ <br/><br/>Two ordered pairs of each element $n\left({R}_2\right)=2 \times 15=30$ elements <br/><br/>Hence $ R_1-R_2=38-30=8$