Think about it as if you have $n$ unique items that you want to divide between $m$ bins/groups of people/etc. - each size $k_1,k_2,\cdots,k_m$. Obviously the first gets $k_1$ items in $\binom{n}{k_1}$ number of ways. Obviously, the second one has to select from ($n-k_1$) items, which is $\binom{n-k_1}{k_2}$ and so on. Now take the product of these binomial coefficients and cancel out some terms to obtain $\binom{n}{k_1 k_2 \cdots k_m} = \frac{n!}{k_1! k_2! \cdots k_m!}$
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 $4!$.
Best Answer
Think about it this way:
Let's line up all ten people, and let's say that the first five go on team 1, and the next five on team 2. Now, there are $10!$ different ways to line up the $10$ people. However, note that you get the same two teams if:
you shuffle the first five people. Having, say, persons 2 and 4 change positions in the line-up will not change the teams: persons 2 and 4 will still be on team 1. Since there are $5!$ ways to shuffle the first five people, you need to divide the original $10!$ by $5!$
you shuffle the persons in positions 6 through 10. Same story. So again, divide by $5!$
You swap the first five people and the second five people. If the teams are indistinct, then the same five people will be on one team, and the same other five on another team. So: divide by $2$ or, what is the same thing: divide by the number of ways we can shuffle the groups, and since there are two groups, that can be done in $2!$ ways. Of course, if the teams are distinct (e.g if the first team is 'the Flyers' and the second 'the Eagles', then swapping the groups of people will make a difference, so in that case do not divide by $2!$