HCF $(x,y) = 16$ and LCM $(x,y) = 48000$. Then the possible number of pairs $(x,y)$ Let $S$ be the set of all ordered pairs $(x,y)$ of positive integers, with HCF $(x,y) = 16$ and LCM $(x,y) = 48000$. The
number of elements in $S$ is
My Attempt : $48000= 2^7. 3 . 5^3$  As the L.c.m contains $2^7$ as a factor and G.c.d contains $2^4$ as a factor , we can assure that one element contains $2^7$ and the other one contains $2^4$. And $3$ and $5^3$ will be divided between two numbers in a manner so that the L.C.M and G.C.D remain as given.
So the possible pairs are $(2^7 .3 . 5^3 , 2^4)$ , $(2^7 .3 , 5^3.2^4)$ , $(2^7 , 3.5^3.2^4)$ , $(2^7.5^3 , 3.2^4)$.
So I think the number of elements in the set $S$ is $4$. Have I gone wrong anywhere? Can anyone please help me ?

WLOG let $\dfrac xX=\dfrac yY=16$ so that $(X,Y)=1$
We have $48000\cdot16=xy=16^2XY\iff XY=3000=3\cdot2^3\cdot5^3$
So, the possible values of $X$ can be 
take none of the factors $$1$$
take one of the factors $$3;2^3;5^3$$
take two of the factors $$2^3\cdot5^3;3\cdot5^3;2^3\cdot3$$
take all three of the factors $$3\cdot2^3\cdot5^3$$
