# How do we get the Statistical weight for derivation of gibbs entropy formula?

I was reading Statistical Physics by F. Mandl. And in Chapter 2.5 pg.61 he derived the general definition of entropy(gibbs entropy forumla). But there is a step that I do not understand. This is the flow of thought:

For a macrostate consisting of an ensemble of a large number, $v$, of identical systems. The number of systems being in state $r$ is $v_r$. Therefore there is the constraint: $$\sum^{N}_{i}v_i=v$$ In total there are $N$ number of states

Mandl goes on to state that the statistical weight $\Omega_v$ of the ensemble when $v_1$ systems are in state 1, $v_2$ systems are in state 2, and so on and so forth all the way to $v_N$ systems are in state $N$..is given by: $$\Omega_v = \frac{v!}{v_{1}!v_{2}!...v_{N}!}$$

This is what I do not understand. If the statistical weight of the systems is the number of ways this distribution of states can be realised, then shouldn't $\Omega_v = \Omega_{v_1}\Omega_{v_2}...\Omega_{v_N}$? i.e. $\Omega_v$ should be given by the following formula: $$\Omega_v=\frac{(v!)^N}{(v_{1}!v_{2}!...v_{N}!)(v-v_{1})!(v-v_{2})!...(v-v_{N})!}$$ since $$\Omega_{v_i} = {v \choose v_i} = \frac{v!}{v_i!(v-v_i)!}$$ Furthermore, even by approximating $v_i << v$, i.e. $N$ is very large, the approximated $\Omega_v$ does not converge to the one written to mandl, instead it converges to $\Omega_v = \frac{v!}{v_1!v_2!...v_N!\cdot v}$.

The $\Omega_v$ are multinomial coefficients. They are a direct extension of the binomial coefficients ${{v}\choose{v_1}}$.
The binomial coefficients count the number of ways to put $v_1$ and $(v-v_1)$ out of $v$ objects into two distinct bins (states).
The multinomials count the number of ways to put $v_1, v_2, \dots v_N$ out of $v$ object ($\sum v_i = v$) into $N$ distinct bins (states). That is exactly what you want to do here.
1. Given $v$ systems. How many ways are there to chose $v_1$ of them to be in state 1? $$\Omega_{1}={{v}\choose{v_1}}$$
2. For the remaining $v-v_1$, how many ways are there to pick $v_2$ of them to be in state 2? $$\Omega_2={{v-v_1}\choose{v_2}}$$
3. Iterate until $v_N$. $$\Omega_v = \prod_i \Omega_i = \frac{v!(v-v_1)!(v-v_1-v_2)!\cdots}{v_1!(v-v_1!)\cdot v_2(v-v1-v2)!\cdots} = \frac{v!}{v_1!v_2!\cdots v_N!}$$