We can also make it work like this. Fix naturals $m_1,...,m_k$ such that $m_1+...+m_k=n$ and let $M$ be a set with $n$ elements. Let $\Omega=\{(A_1,...,A_k)\colon\text{partition of $M$},|A_i|=m_i,\,i=1,...,n\}$. Then $|\Omega|$ is the integer we are looking for and we need to show it's the multinomial coefficient ${n\choose m_1,...,m_k}$. Now let $\Omega^{\prime}=\{(w_1,...,w_n)\in M^n\colon w_i\neq w_j\text{ for }i\neq j\}$. Clearly, $|\Omega^{\prime}|=n!$.
Each element of $\Omega^{\prime}$, i.e. each ordering of $M$, induces an element of $\Omega$, i.e. a partition of $M$, by taking $A_1$…