On counting balls in an urn, combinatorics Suppose you have an urn with $N$ balls where $r$ balls are red and $N-r$ are blue. The balls are identical. Suppose we blindly withdraw at once $k$ balls where $k < r $ and $k < N-r$.
Our outcomes are of the form $ (a_1,a_2,....,a_k )$ where $a_i$ is $r$ or $b$ so for instance one such outcome can be $(b,b,b,r,r,r,b,b,b,...,b)$.
If we want to have say $10$ blue balls and $k-10$ red balls, then we can count $\dfrac{ k! }{10! (k-10)!}$ possible outcomes which is just ${k \choose 10}$
However, if the balls werent identicals, then we can enumerate them. This time we would have ${N-r \choose 10} \cdot {N \choose k-10} $ possible outcomes.
IS this correct? Isnt this just the same as finding number of heads in $N$ tosses of a coin? at the part where the balls are identical.
Thanks in advanced!
 A: When you say that the balls are withdrawn all at once but then describe the outcomes in terms of sequences of colors, I understand you to mean that they are drawn sequentially but without replacement.
If you draw $10$ blue balls and $k-10$ red balls, and they are distinguishable only by color, there are, as you say, $\binom{k}{10}$ distinguishable sequences in which they can be drawn, since there are $\binom{k}{10}$ possible choices for the positions of the $10$ blue balls in the sequence.
That’s fine if you just want to know how many distinguishable sequences of $10$ blue and $k-10$ red balls are possible, irrespective of how many other balls are left in the urn.
If the balls are individually identifiable — e.g., if they are numbered — then there are $\binom{N-r}{10}$ possible sets of $10$ blue balls and $\binom{r}{k-10}$ possible sets of $k-10$ red balls, so there are $\binom{N-r}{10}\binom{r}{k-10}$ different sets of $k$ balls comprising $10$ blue and $k-10$ red balls. That is not, however, the number of possible outcomes, assuming that we are still drawing the balls sequentially without replacement: each of those sets of $k$ balls can be drawn in $k!$ different orders, and each order is a different outcome, so there are actually
$$k!\binom{N-r}{10}\binom{r}{k-10}$$
different outcomes in this case.
If you really are drawing the balls all at once, not in an ordered sequence, then there are of course $\binom{N}k$ different sets of $k$ balls that you could draw, and $\binom{N-r}{10}\binom{r}{k-10}$ of them have $10$ blue and $k-10$ red balls; that is true irrespective of whether the balls are individually identifiable. Once you have one of those $\binom{N-r}{10}\binom{r}{k-10}$ sets, you could ask in how many distinguishable ways it could be lined up. If the balls are distinguishable only by color, the answer is $\binom{k}{10}$: that’s the first problem all over again. If they are individually distinguishable, the answer is $k!$, and the colors don’t matter.
A: You have be clear of several things and define them well.

*

*The $k$ balls which you withdraw - do you treat this as just a set (i.e. unordered collection) of $k$ elements or do you treat it as an ordered collection of balls (i.e. a sequence of $k$ elements)?


*The balls are always identical in touch (I assume) i.e. when you poke in the urn you cannot say which one is which (by touch), otherwise you would run into a completely different set of complications. But... are the balls distinguishable or not once you withdraw them? E.g. if all red balls are numbered uniquely ($1$ to $r$) and all blue balls too (e.g. $1$ to $N-r$), then you have one problem. If you have no numbers on the balls you have a different problem.
Based on the answers of 1) and 2) you can have up to 4 different problems. So you have to be very clear which one you're asking about and trying to solve.
P.S. The notation you used $(b,b,b,r,r,r,b,b,b,...,b)$ somehow implies you have an ordered collection (answer to q.1) of indistinguishable balls (answer to q.2). But as I said you can have different variants of this problem based on the answers of questions 1) and 2).
A: Not quite, you have chosen $\binom{N-r}{10}$ the blue balls, and you have to choose the $\binom{r}{k-10}$ red balls and then you have to shuffle them on how did you draw them so
$$\binom{r}{k-10}\binom{N-r}{10}\cdot k!$$
