Firmi-Dirac Stats. - Calculate number of microstates.

[eric2921]

Homework Statement

Determine the number of microstates where two particles have have energy 2eV, three have 3eV, three have energy 4eV, and two have 5eV.

10 indistinguishable Particles
1 particle allowed per state.

ε₁ = 1eV ε₂ = 2eV ε₃ = 3eV ε₄ = 4eV ε₅ = 5eV ε₆ = 6eV
g₁ = 1 g₂ = 5 g₃ = 10 g₄ = 10 g₅ = 5 g₆ = 1

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Homework Equations

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possibly $\overline{n}$_i=$\frac{1}{\textbf{e}^{(ε_{i}-μ)/kT}+1}$

The Attempt at a Solution

Since the particles are indistinguishable I thought that there would only be one microstate, but I don't think that this is right... can someone point me in the right direction on how to start this problem?

 

[eric2921]

figured it out, was missing a very relevant equation.
$w_{\text{FD}}\left(N_1,N_2,\text{...}N_n\right)=∏ \frac{g_j!}{N_j!\left(g_j-N_j\right)!}$
$w_{\text{FD}}\left(N_1,N_2,N_3,N_4,N_5,N_6\right)=\left(\frac{g_1!}{N_1!\left(g_1-N_1\right)!}\right)\left(\frac{g_2!}{N_2!\left(g_2-N_2\right)!}\right)\left(\frac{g_3!}{N_3!\left(g_3-N_3\right)!}\right)\left(\frac{g_4!}{N_4!\left(g_4-N_4\right)!}\right)\left(\frac{g_5!}{N_5!\left(g_5-N_5\right)!}\right)\left(\frac{g_6!}{N_6!\left(g_6-N_6\right)!}\right)$
$w_{\text{FD}}(0,2,3,3,2,0)=\left(\frac{1!}{0!(1-0)!}\right)\left(\frac{5!}{2!(5-2)!}\right)\left(\frac{10!}{3!(10-3)!}\right)\left(\frac{10!}{3!(10-3)!}\right)\left(\frac{5!}{2!(5-2)!}\right)\left(\frac{1!}{0!(1-0)!}\right)$
evaluates to 1,440,000