I have urn with 5 white balls and 4 black balls.
The balls are non-distinct ( there aren't numbers on it, but every ball have one color: white or black, so they are a little distinct - there is difference between black and white ball ).
In how many ways I can take from urn 3 balls, which are white? I take all balls in the same moment.
When I have ditinct balls, this exercise is very simple. For example white balls have numbers ( 1-5 ) on it. I can create 10 combinations for example {1,2,3} , {2,3,4}. So this is 5 choose 3.
But what if I have non-distinct balls? For me I have only one way to do this. When I take 3 balls and all balls are white, next I return to urn all balls and repeat process and I get 3 white balls and I don't know they are the same, which were in first draw. All white balls are non-distinct, so In every draw when I get 3 white balls I will have { White, White, White }.
Am I right?
I believe I understand what the question is asking.
You could in a blind draw of $3$ balls have drawn $(0W,3B), (1W,2B), (2W,1B)\; or\; (3W,0W)$
These are all the possibilities.
If you add up $\binom50\binom43 + \binom51\binom42 +.... \binom53\binom40$, the answer will come out to be $84$, which is ( and has to be) the same as $\binom93$
So in this context, the question is asking how many ways can you draw all $3$ as white, for which the answer is $\binom53\binom40 = \binom53$
The answer for drawing $3$ black balls is similarly $\binom43$ and so on.
Note:
Distinguishability only by color and a blind draw of all three balls simultaneously is enough to give these results.