3300 = 2×2×3×5×5×11

3300 has 2 factors each of 2 and 5, and 1 factor each of 3 and 11

So for p and q to have LCM = 3300, the following must be true:

A. One or the other of p and q must have 2 factors of 2 and the
other have 0,1, or 2 factors of 2.  The same must be true about
factors of 5.  

Also,

B. One or the other of p and q must have 1 factor of 3 and the
other have 0 or 1 factor of 3.  The same can be said about
factors of 11.
 
So let's list the cases:

Factors of 2:

Case 1: both p and q have 2 factors of 2.
Case 2: p has 2 factors of 2 and q has 0 factors of 2.
Case 3: p has 2 factors of 2 and q has 1 factor  of 2.
Case 4: p has 0 factors of 2 and q has 2 factors of 2.
Case 5: p has 1 factor  of 2 and q has 2 factors of 2.

That's 5 ways the factors of 2 can be had by p and q.

Factors of 3:

Case 1: both p and q have 1 factor of 3.
Case 2: p has 1 factor  of 3 and q has 0 factors of 3.
Case 3: p has 0 factors of 3 and q has 1 factor  of 3.

That's 3 ways the factors of 3 can be had by p and q.

Factors of 5: Same as for factors of 2
That's 5 ways the factors of 5 can be had by p and q.

Factors of 11: Same as for factors of 11
That's 3 ways the factors of 5 can be had by p and q.

Answer: 5×3×5×3 = 225

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Note: The 225 counts, for instance, (44,75) and (75,44) as two
different couples.  Suppose you want to avoid such duplicates
as that and you want to count only the cases where say p≦q,
Then since the only case of p=q is (3300,3300), we can subtract
that case and divide by 2.  That is 225-1=224, and then divide
that by 2, getting 112.  Then we add the couple (3300,3300) that
we subtracted, and that would give 113 couples where p≦q.

Edwin