If ordinary water molecules can contain any of the three hydrogen isotopes, $\ce{^1H}$, $\ce{^2H}$, $\ce{^3H}$, and any of the three oxygen isotopes, $\ce{^{16}O}$, $\ce{^{17}O}$, $\ce{^{18}O}$, how many kinds of water molecules $\ce{H2O}$ are there?

I'm a little confused because since there are $\binom{3}{2}$ ways of choosing two hydrogen atoms and 3 ways of choosing oxygen atoms, shouldn't there be 9 different kinds of water molecule? Why would it be 18 (as the answer says)? In other words, why would the order of hydrogen atoms matter here?

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    $\begingroup$ 3 choose 2 isn't correct here, as that neglects the possibility that you can have the same isotope of hydrogen twice. $\endgroup$ Sep 5, 2017 at 14:06

1 Answer 1


In an example as small as this, you could hypothetically construct all the cases, or at least enough of them to see that $9$ is incorrect.

The only thing you need to change to get the right answer is instead of using $3\choose{2}$ which is a combination without replacement, you need to do combinations with replacement so ${2+3-1}\choose{2}$$=6$ ways to choose the hydrogens. We can very quickly verify this result by writing out the pairs of masses and seeing how many unique combinations we can make: 11 22 33 12 13 23. And then of course each of those combinations can have one of the three different types of oxygen, so 18 possibilities total. If you haven't seen combinations with replacement before, go here to learn more about them.


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