I am confused regarding the following problem in combinatorics ( statistical mechanics ).
Suppose I have the following relation : $$\sum_{i=1}^N n_i=\bar{N}$$
I have to find out the number of possible distribution sets, that satisfy this. For example, suppose, $n_1+n_2=\bar{N}=5.$ This is basically the number of partitions of $5$ into two elements. There are $6$ ways of doing this : (0,5);(5,0);(1,4);(4,1);(2,3);(3,2).
I'm told that this is equivalent to the number of ways of distributing the constant $\bar{N}$ among $N$ elements. Using star and bar (bose-einstein counting), this is equivalent to :
$$\Omega=\frac{(\bar{N}+N-1)!}{\bar{N}!(N-1)!}$$
So, this is comparable to the number of ways of putting $\bar{N}$ balls into $N$ boxes. Since these are just numbers, we are considering identical balls. For example $(n_1,n_2)=(0,5)$ can be thought of as $0$ balls in box $n_1$ and $5$ identical balls (1's) in box $n_2$.
But now suppose, we have distinguishable balls. So, now I want to know the number of ways of putting $\bar{N}$ distinguishable balls into $N$ boxes.
Suppose there is a combination set {$n_i$}$=${$n_1,n_2,n_3,.....,n_N$} that satisfies the above constraint. The number of ways of attaining this configuration would be given by :
$$\omega_i=\frac{\bar{N}!}{n_1!n_2!...n_N!}$$
Now, I need to sum over all possible configuration sets {$n_i$}'s. This is given by : $$\sum_{\{n_i\}} \omega{\{n_i\}}$$
That would give me the total number of possible configurations for distinguishable balls.
In that case this summation would have exactly $\Omega=\frac{(\bar{N}+N-1)!}{\bar{N}!(N-1)!}$ terms in it, as $\Omega$ is the number of possible arrangements for the identical particles.
However, I also know that the total number of possible arrangements for $\bar{N} $ distinguishable balls into $N$ boxes would be $N^{\bar{N}}$.$\Omega$ is the number of distribution sets, that satisfy the constraint $\sum_{i=1}^{N}n_i=\bar{N}$.
So can I then claim,
$$\sum_{\{n_i\}} \omega{\{n_i\}} =\sum_{j=1}^{\frac{(\bar{N}+N-1)!}{\bar{N}!(N-1)!}} \omega_j=N^{\bar{N}}$$
Is this claim correct ?
For example, in identical balls case, $\omega_i=1$ for all {$n_i$}'s and so, total number of possible configurations would be simply $\sum_{i=1}^{\Omega} 1 = \Omega=\frac{(N+\bar{N}-1)!}{\bar{N}!(N-1)!}$
You are correct. This is a consequence of the multinomial theorem, which states that for all complex numbers $a_1,\dots,a_m$, that $$ (a_1+\dots+a_m)^N=\sum_{\substack{k_1+\dots+k_m=n \\ k_1,\,\dots,\,k_m\ge 0}}\frac{N!}{k_1!\cdots k_m!}a_1^{k_1}\cdots a_m^{k_m} $$ In particular, if you set $a_1\gets 1,a_2\gets 1,\dots, a_m\gets 1$, then the LHS is $m^N$, while the RHS is the sum of $N!/(k_1!\cdots k_m!)$ over all possible choices of the integer vector $(k_1,\dots,k_m)$.
This is exactly analogous to how you can prove the sum of all binomial coefficients $\binom{N}{k}$ for a fixed $N$ is equal to $2^N$, by taking the binomial theorem $(x+y)^N=\sum_{k=0}^n \binom{N}kx^ky^{n-k}$, and setting $x\gets 1,y\gets 1$.