Say we have $N$ cubic spaces and $k<N$ particles which we want to assign to these cubes. We can group these into three: protons, electrons and neutrons. Assuming $n_{e^-}\approx n_{p^+}$ and $n_{\text{neutrons}}\approx n_{p^+}/8$ and by $n_{p^+}+n_{e^-}+n_{\text{neutrons}}=k$, we get that $n_{p^+}=n_{e^-}=8k/17$ and $n_{\text{neutrons}}=k/17$. My question is how may I calculate the number of possible arrangements or distributions of these particles in all these cubes?

This is my approach:

I'm not sure if the order in which we position each type of particle matters, but I'll do protons, electrons and neutrons:

The number of possible arrangements for protons would be $$\binom{N}{\frac{8k}{17}},$$

for electrons having placed all protons, $$\binom{N-\frac{8k}{17}}{\frac{8k}{17}},$$

and for the neutrons having placed all the others, it'd be $$\binom{N-\frac{16k}{17}}{\frac{k}{17}}.$$

Finally, we then multiply each of these binomial coefficients:

$$ \begin{aligned} \binom{N}{\frac{8k}{17}}\binom{N-\frac{8k}{17}}{\frac{8k}{17}}\binom{N-\frac{16k}{17}}{\frac{k}{17}}&=\dfrac{N!}{\left(\frac{8k}{17}\right)!(N-\frac{8k}{17})!}\dfrac{(N-\frac{8k}{17})!}{\left(\frac{8k}{17}\right)!(N-\frac{16k}{17})!}\dfrac{(N-\frac{16k}{17})!}{\left(\frac{k}{17}\right)!(N-k)!}\\ &=\dfrac{N!}{\left(\dfrac{k}{17}\right)!\left(\left(\dfrac{8k}{17}\right)!\right)^2(N-k)!}. \end{aligned}$$

  • $\begingroup$ Assuming only one particle can go in each space, your answer is correct. $\endgroup$ Sep 19 at 16:55
  • $\begingroup$ Thanks for confirming Mike. $\endgroup$ Sep 19 at 18:04


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