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I'm struggling with applying my counting skills to probability problems. Specifically I'm grappling with how to enumerate the number of ways to draw colored balls from an urn in which we may have some balls different colors than others.

Here is an example.

Suppose we have $6$ red balls and $4$ blue balls and we want to know the number of ways to remove $4$ of them without replacement in order to calculate our sample space. The answer to me would be $10 \choose 4$. But it seems in most probability problems involving removing balls from urns the actual appropriate sample space would require us to "label" the colored balls so the actual sample space would be $10\cdot 9 \cdot 8 \cdot 7$.

If given a problem where we are removing balls from an urn and we are concerned with specific outcomes, for say the $n$th ball, is it necessary to view the balls within their own groups as labeled? That is, does the fact that we are removing balls and concerned with what balls gets removed when impose an ordering on our objects?

For instance, would it be better me to think of a probability question involving $6$ red balls and $4$ blue balls as the set $\{R,R,R,R,R,R,B,B,B,B\}$? Or the set $\{R_1,R_2,R_3,R_4,R_5,R_6,B_1,B_2,B_3,B_4\}$ where $R_i \neq R_j$ for $j\neq i$ and $B_i \neq B_j$ for $j\neq i$.

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  • $\begingroup$ $\{R,R,R,R,R,R,B,B,B\}$ is not a set. $\endgroup$ – zhoraster Jan 23 '17 at 14:56
  • $\begingroup$ @zhoraster it is a multiset en.wikipedia.org/wiki/Multiset $\endgroup$ – ClownInTheMoon Jan 23 '17 at 14:57
  • $\begingroup$ Thanks for the reference, but somehow I know what the multiset is. However, there is no word "multiset" in your post. $\endgroup$ – zhoraster Jan 23 '17 at 15:38
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$\binom{10}4$ counts the ways to select 4 distinct items from a set of 10 into a set.   Order of the result doesn't matter.   This may also be written as $^{10}\mathrm C_4$ .

Example: The probability of selecting two red balls among the four selected is: $$\mathsf P(R=2)=\dfrac{\binom 6 2\binom 4 2}{\binom {10}4}$$

The numerator counts ways to select two from six red, and two from the four blue balls.   The denominator counts the ways to select any four from all ten balls.


$\binom{10}4 4!$, or $10\cdotp 9\cdotp 8\cdotp 7$, counts the ways to select 4 distinct items from a set of 10 into a list.   The order of the result matters.   Also written as $^{10}\mathrm P_4$ .

Example: The probability that the third ball is red when drawing four balls is:

$$\mathsf P(X_3=r) = \dfrac{6\cdot{^9\mathrm P_3}}{^{10}\mathrm P_4}$$

Which is, of course, $3/5$, and can be found much easier; but here we demonstrate the principle that the numerator counts ways to select the third ball from 6 red, and select and arrange three other balls from the nine remaining.   In the denominator, then, we must count the ways to select and arrange any 4 from 10 balls.


The main point: "As above, so below."   When order matters in the denominator, it matters in the numerator, and so on.

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If things are same as we have red balls. Then after picking we don't need to take care about the arrangements of balls.

But if we are picking 4 balls R1, R2, R3, R4. We have to take care about their arrangements also.

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It depends if you can distinguish them. If the problem doesn't say it (or it can´t be deduced from the nature of it), both way are right. You just have to be consistent, I mean, if you use the first one, from then and until the end you can't distinguish them, and else you can do it.

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