Let us take a much smaller example, $4$ people. We want to divide them into two groups of two each, one group to wear nice blue uniforms, the other to wear brown and yellow stripes. How many ways are there to do the division? We need to choose who will wear the blue uniforms. This can be done in $\binom{4}{2}=6$ ways.
Now consider the ways to divide them into two groups of two each, no uniforms. Call the people $a$, $b$, $c$, and $d$. As soon as we decide who goes with $a$, we will have done the division. So there are $3$ ways to do the job.
Another way of thinking about the second problem is that we first divide the people into two groups-with-uniform. Then we take away the uniforms. The two old divisions $a$ and $c$ wear blue, $b$ and $d$ wear browm/yellow stripes and $a$ and $c$ wear brown/yellow stripes, and $b$ and $d$ wear blue now become a single division into two groups. So to count the number of divisions into uniormless groups, we divide $\binom{4}{2}$ by $2$.
Remark: The idea generalizes. For example, take $20$ people, and divide them into $4$ groups of $5$ people each, one group to wear blue, another white, another red, and another black. First choose who will wear blue. This can be done in $\binom{20}{5}$ ways. For every way of doing this, there are $\binom{15}{5}$ ways to decide who will wear whie, and then $\binom{10}{5}$ ways to decide who will wear white, for a total pf $\binom{20}{5}\binom{15}{5}\binom{10}{5}$.
This can also be written as the multinomial coefficient $\binom{20}{5,5,5,5}$.
Now how many ways are there to divide them into $4$ uniformless groups? When people take off their uniforms, $4!$ uniformed divisions collapse into one, so we need to divide by $4!$.
Since bins with the same size are indistinguishable we assume ordered arrangements
\begin{align*}
1\leq n_1\leq n_2\leq \ldots \leq n_k\leq n
\end{align*}
with
\begin{align*}
n_1+n_2+\cdots+n_k=n
\end{align*}
and classify them in tuples according to equal bin sizes:
Let $(c_1,c_2,\ldots,c_j)$ be a $j$-tuple with
\begin{align*}
&n_1=n_2=\ldots=n_{c_1}\\
&\qquad <n_{c_1+1}=n_{c_1+2}=\ldots=n_{c_2}\\
&\qquad <\ldots\\
&\qquad <n_{c_{j-1}+1}=n_{c_{j-1}+2}=\ldots=n_{c_j}\qquad\qquad 1\leq j\leq k
\end{align*}
with $n_{c_1}=n_1$ and $n_{c_j}=n_k$.
We conclude
The number of different assignments is
\begin{align*}
\frac{n!}{n_1!\cdots n_k!\cdot c_1!\cdots c_j!}=\frac{n!}{\left(n_{c_1}!\right)^{c_1}\cdots\left(n_{c_j}!\right)^{c_j}\cdot c_1!\cdots c_j!}
\end{align*}
Reasoning:
There are $n!$ different arrangements of $x_1,\ldots,x_n$.
The order of $n_j$ ($1\leq j\leq k$) elements in a bin does not matter, so we divide by $n_j!$.
The order of $c_l$ bins ($1\leq l\leq j$) with equal size $n_{c_l}$ does not matter, so we divide by $c_l!$.
OPs first example with $n=4, k=2, n_1=1, n_2=3$ implies $c_1=c_2=1$ and we obtain
\begin{align*}
\frac{n!}{n_1!n_2!c_1!c_2!}=\frac{4!}{1!3!1!1!}=4
\end{align*}
OPs second example with $n=4, k=2, n_1=2, n_2=2$ implies $c_1=2$ and we obtain
\begin{align*}
\frac{n!}{n_1!n_2!c_1!}=\frac{4!}{2!2!2!}=3
\end{align*}
Best Answer
(1) Here, we have $\displaystyle \binom{9}{4} = \dfrac{9!}{4!5!}$. The point that matters here is that it is only after one child we'll call "A" takes a stance as a self-appointed captain to pick the rest of his/her team that the groups become distinguishable: A's group, and A's rejects. So Captain "A", has in fact, nine-choose $4$ ways to select the other four children on his/her now-identifiable group: the "A-team". The five unpicked (A-rejected) children become, by default, A-team's competition.
(2) We can use either the binomial coefficient or its equivalent expression as a multinomial coefficient: $\displaystyle \binom{10}{x_1} = \binom{10}{x_1, x_2} = \dfrac{10!}{x_1!\cdot x_2 !}$, because $x_1 + x_2 = 10$.