If r, s, t are prime numbers and p, q are the positive integers such that the LCM of p, q is , then the number of ordered pair (p, q) is-

The correct answer is: 225

Step by step solution :
It is given that 
That is, at least one of $p$ and $q$ must have ,  and  in their prime factorizations.
Now, consider the cases for power of $r$ as follows:
Case 1 : p contains   then q has   with K = (0,1,2,3)
That is, number of ways$=4$.
Case 2 : q contains   then q has   with K = (0,1,2,3)
That is, number of ways$=4$.
Case 3 : Both p and q contains  
Then, number of ways$=1$.
Therefore, exponent of $r$ may be chosen in $2+2+1=5$ ways.
Similarly, exponent of s may be chosen in $4+4+1=9$ ways and 
Exponent of $s$ may be chosen in $2+2+1=5$ ways
Thus, the total number of ways is:
 
Hence, the number of the ordered pair $(p,q)$ is $225$.

Be careful about selecting the r s and t for this particular problem. Because number ways for each selection of r, s and t there are 3 cases. Because there are two values given