Let R be a relation from the set \ 1, 2, 3 , 60\ to itself such...

Given $b = pq$ where $p \text{\text{ and } }$ 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 $p = 5q = 5, 7, 11$ 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, ⋯ , 60\right\}$</p><p>i.e. total$60$possibilities</p><p>Hence, the number of relations$= 60 \times 11 = 660