How many ways to realize a set of population?

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SUMMARY

The discussion focuses on calculating the number of ways to realize a set of populations for three distinguishable particles A, B, and C, with specified energy levels. The Boltzmann formula W(𝑛) = N! ∏(g_α^n_α/n_α) is applied to determine the configurations, leading to a conclusion of three distinct arrangements: (A, BC), (A, CB), and (B, C, A). The user initially miscalculated the configurations but later recognized their errors, confirming the correct application of statistical mechanics principles.

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teddd
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Hi everyone!

Here's my problem of the day:

Let's take a box containing 3 identical (but distinguishable) particles A B C. Let this be a canonical ensamble.

Suppose that A has energy [itex]\varepsilon_0[/itex] and both B and C have energy [itex]\varepsilon_1[/itex]. We thereforre have 2 energy level, [itex]n_0,n_1[/itex]. Take the number of states [itex]g_{\alpha}[/itex] in each energy level [itex]\varepsilon_{\alpha}[/itex] to be [itex]1[/itex].Now, I want to calculate in how many ways the set of population [itex]\vec{n}=(n_0,n_1)[/itex] can be realized.

At first sight I'd say that they're two: I can take [itex](A,BC)[/itex] or [itex](A,CB)[/itex], being the particle distinguishable.

But if I use the well-known Boltzmann forumula [tex]W(\vec{n})=N!\prod_{\alpha}\frac{ g_{\alpha}^{n_{\alpha}}}{n_{\alpha}}[/tex] and I put in the g's and n's I've taken above I get:[tex]W(\vec{n})=3! \left(\frac{1^1}{1!}\frac{1^2}{2!}\right)=3[/tex]so there should be three ways to set up the vector [itex]\vec{n}[/itex]!Where am I mistaking?? Thanks for help!
 
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Ah, well, it's ok! I figured that out, I've done some serious rookie mistakes...
 

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