For an Einstein solid with each of the following values of N and q , list all of the possible microstates, count them, and verify formula $\mathrm{\Omega }(N,q)=\left(\begin{array}{c}q+N-1\\ q\end{array}\right)=\frac{(q+N-1)!}{q!(N-1)!}$

(a) $N=3,q=4$

(b)$N=3,q=5$

(c) $N=3,q=6$

(d) $N=4,q=2$

(e) $N=4,q=3$

(f) $N=1,q=$anything

(g) N= anything, $q=1$

(a)We can represent the energy unit with (•) and the partition with (mid): So, let's say we have three oscillators$N=3$and four units of energy$q=4$, the possible microstates are:

Using the general formula for Einstein solid multiplicity with $N=3$and$q=4$

$\begin{array}{c}\mathrm{\Omega }(N,q)=\left(\begin{array}{c}q+N-1\\ q\end{array}\right)=\frac{(q+N-1)!}{q!(N-1)!}\\ \mathrm{\Omega }(3,4)=\left(\begin{array}{c}4+3-1\\ 4\end{array}\right)=\frac{(4+3-1)!}{4!(3-1)!}=15\end{array}$

(b) For three oscillators$N=3$and five units of energy $q=5$, the possible microstates are:

Using the general formula for Einstein solid multiplicity with $N=3$and$q=5$:

$\begin{array}{c}\mathrm{\Omega }(N,q)=\left(\begin{array}{c}q+N-1\\ q\end{array}\right)=\frac{(q+N-1)!}{q!(N-1)!}\\ \mathrm{\Omega }(3,5)=\left(\begin{array}{c}5+3-1\\ 5\end{array}\right)=\frac{(5+3-1)!}{5!(3-1)!}=21\end{array}$

(c) For three oscillators $N=3$and five units of energy $q=6$, the possible microstates are:

Using the general formula for Einstein solid multiplicity with $N=3$and $a=6$:

$\begin{array}{c}\mathrm{\Omega }(N,q)=\left(\begin{array}{c}q+N-1\\ q\end{array}\right)=\frac{(q+N-1)!}{q!(N-1)!}\\ \mathrm{\Omega }(3,6)=\left(\begin{array}{c}6+3-1\\ 6\end{array}\right)=\frac{(6+3-1)!}{6!(3-1)!}=28\end{array}$

(d) For four oscillators$N=4$and two units of energy$q=2$the possible microstates are:

Using the general formula for Einstein solid multiplicity with $N=4$and $q=2$

$\begin{array}{c}\mathrm{\Omega }(N,q)=\left(\begin{array}{c}q+N-1\\ q\end{array}\right)=\frac{(q+N-1)!}{q!(N-1)!}\\ \mathrm{\Omega }(4,2)=\left(\begin{array}{c}4+2-1\\ 2\end{array}\right)=\frac{(4+2-1)!}{2!(4-1)!}=10\end{array}$

(e) For four oscillators $N=4$and three units of energy$q=3$the possible microstates are:

Using the general formula for Einstein solid multiplicity with$N=4$and$q=3$

$\begin{array}{c}\mathrm{\Omega }(N,q)=\left(\begin{array}{c}q+N-1\\ q\end{array}\right)=\frac{(q+N-1)!}{q!(N-1)!}\\ \mathrm{\Omega }(4,3)=\left(\begin{array}{c}4+3-1\\ 3\end{array}\right)=\frac{(4+3-1)!}{3!(4-1)!}=20\end{array}$

(f)Using the general multiplicity of Einstein solid formula for N oscillators$N=1$and q units of energy $q=q$:

$\mathrm{\Omega }(N,q)=\left(\begin{array}{c}q+N-1\\ q\end{array}\right)=\frac{(q+N-1)!}{q!(N-1)!}$

$\mathrm{\Omega }(1,q)=\left(\begin{array}{c}1+q-1\\ q\end{array}\right)=\frac{(1+q-1)!}{q!(1-1)!}=\frac{q!}{q!}$

$\mathrm{\Omega }(1,q)=1$

(g)Using the general multiplicity of Einstein solid formula for N oscillators)$N=N$and one units of energy $q=1$:

$\mathrm{\Omega }(N,q)=\left(\begin{array}{c}q+N-1\\ q\end{array}\right)=\frac{(q+N-1)!}{q!(N-1)!}$

$\mathrm{\Omega }(N,1)=\left(\begin{array}{c}N+1-1\\ 1\end{array}\right)=\frac{(N+1-1)!}{1!(N-1)!}=\frac{N(N-1)!}{(N-1)!}$

$\mathrm{\Omega }(N,1)=N$