Possible orderings when the items are not unique? First of all, I'm sure this question has been answered somewhere on the web, but I am just starting probability and I don't have the vocabulary to know what to look for, which is why I am asking here.
I have $5$ white tokens and $5$ black tokens and I want to find the the number of possible orderings.
I know how to do this normally, it would be $10!$, but here I don't care about the specific order of things, just the overall pattern of black and white.
In other words $B_1 W_1 B_2 W_2$ is the same as $B_2 W_1 B_1 W_2$.
I was thinking I could take that $10!$ and somehow remove the different options for white and black. Something like $\frac{10!}{5!5!}$  because there are $5!$ orders of white and $5!$ orders for black. But I have no idea if this is correct, and I'm not sure how I would go about testing it.
 A: Your answer is correct.
You have ten positions to fill.  You can fill five of them with white tokens in $\binom{10}{5}$ ways.  You can fill the remaining five of them with black tokens in $\binom{5}{5} = 1$ way.  Hence, the number of ways you can arrange the tokens is 
$$\binom{10}{5}\binom{5}{5} = \binom{10}{5}$$
If you instead had five red tokens, three blue tokens, and two green tokens, you could fill five of the ten positions with red tokens in $\binom{10}{5}$ ways, three of the remaining five positions with blue tokens in $\binom{5}{2}$ ways, and both of the final two positions with green tokens in $\binom{2}{2}$ ways.  Thus, the tokens could be arranged in 
$$\binom{10}{5}\binom{5}{3}\binom{2}{2} = \frac{10!}{5!5!} \cdot \frac{5!}{3!2!} \cdot \frac{2!}{2!0!} = \frac{10!}{5!3!2!}$$
ways.
A: In general, if you have $k_1$ items of type 1, $k_2$ of type 2, ..., $k_n$ items of type $n$, the number of orderings is given by the multinomial coefficient:
$\begin{align}
  \binom{k_1 + k_2 + \dotsb + k_n}{k_1, k_2, \dotsc, k_n}
    = \frac{(k_1 + k_2 + \dotsb + k_n)!}{k_1! \, k_2! \, \dots \, k_n!}
\end{align}$
To prove this, take each of the items as different (i.e., number each item in its type). Then there are $(k_1 + k_2 + \dotsb + k_n)!$ orderings. But if you erase the numbers on the items of type 1, $k_1!$ of the orderings become the same. You can do the same for the others.
In your case, this gives:
$\begin{align}
   \binom{5 + 5}{5, 5} = \binom{10}{5} = 252
\end{align}$
