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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?

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    $\begingroup$ I think Mathematics SE they will have a formula at hands without tediously manually removing the multiple items $\endgroup$ – Alchimista Apr 1 '19 at 12:25
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    $\begingroup$ 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.) $\endgroup$ – Zhe Apr 1 '19 at 18:08
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There are 5 ways to choose the central atom.

Choosing the possible combinations of the other 4 items from 6 isotopes is complicated as this article explains. But it provides a formula for combinations with repetition that gives the answer of 126.

This implies that there are 630 distinct isotope combinations in the molecule (assuming that we can distinguish combinations of isotopes with the same atomic number which is possible with high resolution mass spectroscopy). If only integer atomic numbers count then the distinguisahble combinations will be lower. But the easiest way to work that out is to do a Monte-Carlo simulation not to look up probability formulae.

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