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Let $A=\{1,2,3, \ldots 20\}$. Let $R_1$ and $R_2$ two relation on $\mathrm{A}$ such that $\mathrm{R}_1=\{(\mathrm{a}, \mathrm{b}): \mathrm{b}$ is divisible by $\mathrm{a}\}$ $\mathrm{R}_2=\{(\mathrm{a}, \mathrm{b}): \mathrm{a}$ is an integral multiple of $\mathrm{b}\}$. Then, number of elements in $R_1-R_2$ is equal to_____.
$44$
$46$
$45$
$40$
Solution
$ \mathrm{n}\left(\mathrm{R}_1\right)=20+10+6+5+4+3+2+2+2 $
$ +2+\underbrace{1+\ldots+1}_{10 \text { times }}$
$\mathrm{n}\left(\mathrm{R}_1\right)=66$
$\mathrm{R}_1 \cap \mathrm{R}_2=\{(1,1),(2,2), \ldots(20,20)\}$
$\mathrm{n}\left(\mathrm{R}_1 \cap \mathrm{R}_2\right)=20$
$\mathrm{n}\left(\mathrm{R}_1-\mathrm{R}_2\right)=\mathrm{n}\left(\mathrm{R}_1\right)-\mathrm{n}\left(\mathrm{R}_1 \cap \mathrm{R}_2\right)$
$=\mathrm{n}\left(\mathrm{R}_1\right)-20$
$=66-20$
$\mathrm{R}_1-\mathrm{R}_2=46 \text { Pair }$
