# Number of stable molecules of artificial molecule XY4 [closed]

Suppose element X has $$5$$ stable isotopes, and element Y has $$6$$ stable isotopes. Find the number of natural molecules, knowing that an X has a charge of $$+4$$ and Y has a charge of $$-1$$.

What I tried was this: for element X we have $$5$$ options, and for each element Y we have $$6$$ options. Therefore the total number of options would be $$5 × 6 × 6 × 6 × 6$$.

But this answer is clearly wrong, as it counts the same type of molecule, multiple items. (AAAB) and (AAABA) count twice which we don't want to happen. (A is an isotope of Y and B is some other isotope of B)

What can I do?

• I think Mathematics SE they will have a formula at hands without tediously manually removing the multiple items – Alchimista Apr 1 '19 at 12:25
• You can't answer this question. What are the valid geometries of the $\ce{XY4}$ molecule? Tetrahedral? Square planar? Distorted tetrahedral? See-saw? Technically, these are all different, and you would need to account for them or rule them out, which requires more information than what is given here. (I'm assuming that $\ce{X}$ is the central atom here. If that's not a valid assumption, then it's even more insane.) – Zhe Apr 1 '19 at 18:08