Let R be a relation from the set $$\{1, 2, 3, \ldots, 60\}$$ to itself such that $$R = \{(a, b) : b = pq$$, where $$p, q \geq 3$$ are prime numbers$$\}$$. Then, the number of elements in R is

We have the relation $$R = \{(a, b) : b = pq, \text{ where } p, q \geq 3 \text{ are prime numbers}\}$$ from the set $$\{1, 2, 3, \ldots, 60\}$$ to itself. We need to find the number of elements in $$R$$.

We first identify all possible values of $$b$$. We need $$b = pq$$ where $$p$$ and $$q$$ are primes with $$p, q \geq 3$$, and $$b \leq 60$$. The primes $$\geq 3$$ are $$3, 5, 7, 11, 13, 17, 19, \ldots$$

We systematically list all products $$pq \leq 60$$ (where $$p \leq q$$):

With $$p = 3$$: $$3 \times 3 = 9$$, $$3 \times 5 = 15$$, $$3 \times 7 = 21$$, $$3 \times 11 = 33$$, $$3 \times 13 = 39$$, $$3 \times 17 = 51$$, $$3 \times 19 = 57$$. (Note $$3 \times 23 = 69 > 60$$.)

With $$p = 5$$: $$5 \times 5 = 25$$, $$5 \times 7 = 35$$, $$5 \times 11 = 55$$. (Note $$5 \times 13 = 65 > 60$$.)

With $$p = 7$$: $$7 \times 7 = 49$$. (Note $$7 \times 11 = 77 > 60$$.)

So the distinct values of $$b$$ are: $$9, 15, 21, 25, 33, 35, 39, 49, 51, 55, 57$$, which gives us 11 possible values.

Now, for each valid $$b$$, the value of $$a$$ can be any element in $$\{1, 2, \ldots, 60\}$$, giving 60 choices.

Hence the total number of elements in $$R$$ is $$11 \times 60 = 660$$.

Hence, the correct answer is Option B.