How many different pairs of whole numbers have HCF 9 and LCM 1188?

Write the numbers as 9$m$ and $9n$ where $m$ and $n$ are co-prime. Then the LCM is $9mn$, so $mn=1188÷9=132=2^2\times 3\times 11$. Because $m$ and $n$ share no factor, each of the three prime blocks (4, 3 and 11) must go wholly to one side, giving $2^3=8$ ordered splits and 8 ÷ 2 = 4 unordered pairs: (1, 132), (3, 44), (4, 33) and (11, 12). The number pairs are 9 & 1188, 27 & 396, 36 & 297 and 99 & 108. Forgetting to halve the ordered count gives 8.