Solution:

Step 1: Finding the number of pairs (p,q) whose LCM is 1800.

We can find the number of pairs (p,q) whose LCM is 1800 by using the following formula:

LCM(p,q) = (p x q)/GCD(p,q)

So, we need to find all pairs (p,q) such that (p x q)/GCD(p,q) = 1800.

Let GCD(p,q) = d. Then, we can write p = dx and q = dy, where x and y are coprime.

Substituting this in the above equation, we get:

dx x dy/d = 1800

=> xy = 1800/d

Now, we need to find all possible values of d for which xy = 1800/d has solutions in coprime integers x and y.

Step 2: Finding the number of ways of expressing k as a product of two coprime numbers.

Let's say k has prime factorization: p1^a1 x p2^a2 x … x pn^an

Then, the number of ways of expressing k as a product of two coprime numbers is given by:

[(2^(n-1)) - (a1 > 0) - (a2 > 0) - … - (an > 0)]

where (a1 > 0) means that a1 is not equal to zero, and so on.

Step 3: Putting it all together.

To find the number of pairs (p,q) whose LCM is 1800, we need to find all possible values of d for which xy = 1800/d has solutions in coprime integers x and y.

The prime factorization of 1800 is: 2^3 x 3^2 x 5^2.

So, we need to find all possible values of d for which xy = 2^3 x 3^2 x 5^2 / d has solutions in coprime integers x and y.

Let's consider each prime factor separately:

For the prime factor 2, we have 4 choices:
- d = 1, then xy = 2^3 x 3^2 x 5^2, which has (3+1)(2+1)(2+1) = 84 solutions in coprime integers x and y.
- d = 2, then xy = 2^2 x 3^2 x 5^2, which has (2+1)(2+1)(2+1) = 27 solutions in coprime integers x and y.
- d = 4, then xy = 2 x 3^2 x 5^2, which has (1+1)(2+1)(2+1) = 24 solutions in coprime integers x and y.
- d = 8, then xy = 3^2 x 5^2, which has (2+1)(2+1) = 9 solutions in coprime integers x and y.

For the prime factor 3, we have 3 choices:
- d = 1, then xy = 2^3 x 5^2

LCM of 1800=2^3.3^2.5^2; n=3{2,3,5}.The no of ways in which a composite number n can be resolved into two factors which are relatively Coprime to each other =2^n-1 =2^3-1=4