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 ?

If $$\gcd(x,y)=16=2^4$$ and $$\operatorname{lcm}(x,y)=48000=2^7\cdot3\cdot5^3$$ then $$xy=2^{11}\cdot3\cdot5^3$$
So we can let 
$$x = 2^\alpha\cdot3^\beta\cdot5^\gamma \qquad \text{and} \qquad
  y = 2^{11-\alpha}\cdot3^{1-\beta}\cdot5^{3-\gamma}$$
We need
$$\begin{align}
   \min(\alpha, 11-\alpha) &= 4 \\
   \max(\alpha, 11-\alpha) &= 7 \\
   \min(\beta, 1-\beta) &= 0 \\
   \max(\beta, 1-\beta) &= 1 \\
   \min(\gamma, 3-\gamma) &= 0 \\
   \max(\gamma, 3-\gamma) &= 3
\end{align}$$
So $$\begin{align}
   \alpha &\in \{4,7\} \\
   \beta &\in \{0,1\} \\
   \gamma &\in \{0,3\}
\end{align}$$
So there are $2\cdot2\cdot2 = 8$ possible $(x,y)$ pairs.
