Is it necessary to apply distinguishability to two types of objects in permutations and combinatorics? Given a one-dimensional array of locations, and 2N indistinguishable beans, N of which are black, and N of which are white, what is the probability that a random distribution of beans will have all the white beans in the left half of the array, and all the black beans in the right half?
I know that the number of ways of obtaining an ordered subset of k elements is $\frac{n!}{(n-k)!}$. So this indicates to me that the number of possible sequences of black and white beans is $\frac{2N!}{(2N-N)!}=\frac{(2N)!}{N!}$. 
I am wondering if the probability of an outcome is one divided by the number of possible sequences. In the cases of N = 1 and N = 2 it is. But, for example, in the case of N = 2, there are four sequences which satisfy the conditions out of a possible 24. $(W_1, W_2, B_1, B_2)$, $(W_1,W_2,B_2,B_1)$, $(W_2,W_1,B_1,B_2)$, $(W_2,W_1,B_2,B_1)$. 
My confusion is coming from not knowing how to apply the indistinguishability of the beans. What is a rigorous way of defining the probability of all white on the left, all black on the right?
 A: Since probability in this case corresponds directly to counting, let's just focus on counting.
Here's a simple example dealing with some indistinguishable things: Rearranging the sequence $AMA$, of which we have only three ways, $AAM, AMA$, and $MAA$.
Naively, one might think that there are $3! = 6$ possibilities, but that would be if we had two distinguishable $A$'s, say $A_1$ and $A_2$. But here, since there were $2! = 2$ ways to rearrange $A_1$ and $A_2$, and those didn't actually affect anything, we have only $3 = \frac{6!}{2!}$ possibilities; we divide by the number of rearrangements that don't affect anything.
To deal with more than one kind of indistinguishable thing, think about rearrangements of $AMMA$, of which there are $6$: $AAMM, AMAM, AMMA, MAAM, MAMA, MMAA$. From the naive but plausible guess of $4! = 24$, we only have $6 = \frac{4!}{2! \cdot 2!}$ distinct rearrangements; dividing by $2!$ once for rearranging the duplicate $A$'s, and then again for the duplicate $M$'s.
A comment on the probability aspect: once we've successfully dealt with indistinguishability and counting total rearrangements, there really is only one way to have all the white beans on the left, isn't there?
