Mathematics - Sets and Relations Question with Solution | TestHub

Given $b=pq$ where $p \& q$ are prime numbers $\ge 3$ and $1\le pq\le 60$

If $p=3,q=3,5,7,11,13,17,19$

And if $p=5 q=5,7,11$ and so on,

So, $b$ will be $3\times 3,3\times 5,3\times 7,3\times 11,3\times 13,3\times 17,3\times 19,5\times 5,5\times 7,5\times 11,7\times 7$

i.e. total $11$ possibilities

Now $a$ can be $\left\{1,2,\cdots ,60\right\}$

i.e. total $60$ possibilities

Hence, the number of relations $=60\times 11=660$