Is Partition Formula wrong when dividing 2 into 2 classes?

Question

In basic combinatorics, when dividing a total of N objects into k classes each with $r_k$ objects we have Partition Formula: $$ \binom{N}{r_1}\binom{N-r_1}{r_2}...\binom{r_k}{r_k}=\frac{N!}{r_1! r_2!...r_k!} $$

But if you try to divide $2$ into $2$ classes, or $3$ into $3$ classes, the formula gives you $2$ and $6$, but apparently there is only one way to divide it. But it functions normally when you divide $2$ into $2$ and zero and $3$ into $1$ and $2$.

Since we deduce the Partition Formula from n choose k method, we can see that in the case of dividing $2$ into $1$ and $1$, it first chooses $1$ out of $2$, which has $2$ ways. And then choose the left one. So this process actually arranges the separation orderly. $(a,b)$ and $(b,a)$ are different. But in the case of dividing $3$ into $1$ and $2$, we end up with $(a, bc)$, $(b, ac)$ and $(c, ab)$. Not ordered, which is good.

Why? I'm confused. Help me.

Answer 1

But if you try to divide 2 into 2 classes, or 3 into 3 classes, the formula gives you 2 and 6, but apparently there is only one way to divide it.

This is the case when the classes are identical, i.e., whether your elements $a$ and $b$ in the first class or in the second class is not important. This formula is given for distinct classes, therefore for "2 into 2" case, we have $2! = 2$ such ways and for "3 into 3" case, we have $3! = 6$ such cases.

But in the case of dividing 3 into 1 and 2, we end up with (a, bc), (b, ac) and (c, ab). Not ordered, which is good.

There is a problem here since formula does not have to count $(bc,a)$, $(ac,b)$, $(ab,c)$ when we divide them with $1+2$ and not $2+1$. So, there are $3$ ways to divide them, which is $\binom{3}{1}\binom{2}{2}$, consistent with the formula.

I think the confusion comes from $1+1+1$ and $1+2$ cases. Notice that when we have $3$ classes for $3$ objects, we can permute the objects into classes since the size of the classes are the same for all three. However, in $1+2$ case, we cannot permute them since it restricts us to have $1$ object in the first class and $2$ objects in the second class.

Answer 2

The multinomial coefficient is $$\left(\begin{array}{c} k\\k_1,\ldots,k_n\end{array}\right) = {k\choose k_1}{k-k_1\choose k_2}\cdots{k-k_1-k_2-\ldots-k_{n-1}\choose k_n},$$ where $k=k_1+\ldots+k_n$ for natural numbers $k_1,\ldots,k_n$, $n\geq 2$. It equals $$\frac{k!}{k_1!\cdots k_n!}.$$ In particular, if $n=2$, then $${k\choose k_1,k_2} = {k\choose k_1}{k-k_1\choose k_1}= {k\choose k_1}{k_2\choose k_2} = {k\choose k_1}$$ and also $${k\choose k_1}={k\choose k-k_1} = {k\choose k_2}.$$