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Look at this problem:

If one is to throw $2$ dice, what is the probability that he/she gets a total of exactly $4$ dots?

Well, the amount of the events that satisfy the condition is $3$ as we have $(1;3)$, $(2;2)$ and $(3;1)$ as three possibilities. The number of all events that can happen is $6^2$ - so we've used the permutation with repetition formula.

Consider another problem:

Out of a box containing $6$ white and $3$ black balls $2$ balls are randomly taken. What is the probability that the balls taken are both black?

This is the solution that is written in a book: the number of all the possible events is $n={9\choose 2}=36$. The amount of events that satisfy the given condition is $m={3\choose 2}=3$. The probability is hence $\frac{1}{12}$.

In this second problem the number of all the possible events is calculated using the combination without repetition formula. Now what I don't understand is this: in the first problem, we have to include both $(1;3)$ and $(3;1)$, while in the second the calculation of the number of the possible events shows that $(W;B)$ and $(B;W)$ should not be both taken into account. Why is it so? Sorry if this is not understandable, I am not a native English speaker.

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You CAN, if you wish, ignore ordering in the dice example, but then the (21) different outcomes are no longer equally likely (the doubles are half as likely as the non-doubles). For that reason, it is easier to model the dice by ordered pairs.

As RandomUser points out, you CAN model the ball problem by ordered pairs but in that case it is just as easy to use unordered pairs (i.e. combinations) and the numbers are smaller, so usually we would do it as you outlined.

When choosing balls with replacement (or rolling dice, or any identical trials) it is often easier to model the experiment with ordered sets (i.e.permutations), even if the events in question don't require it, because we retain the "equally likely outcome" property which makes the computations simpler.

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In the first instance, either die can be any value from $1$ to $6$, giving the $6^2$ ways. Because either die can be any of those, $(3,1)$ and $(1,3)$ have their own probabilities and are considered separate events that add into your total.

The second instance considers any two particular drawn balls as one possible outcome, regardless of what order you draw them in. So you have $9$ possibilities for the first, then $8$ for the second. This would give $72$ if you cared about order, but you don't. There are $2$ ways to arrange them, and $\frac{72}{2}=36$ ways. The choice of $m$ follows similar logic.

Note that you can work out the whole second scenario with order mattering and it still works.

As I said above, with order mattering it's $72$ ways. Then you need two black balls. There are $3\cdot2=6$ ways for that, caring about order. $$\frac{6}{72}=\frac{1}{12}$$

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The distinction is that the results of individual die rolls are independent so we use permutations with repetition to count the microstates for a given macrostate (the sum of the dots). We cannot use combinations because some of the microstates are paired, and some are not.

However, while you can draw the same colour twice, you cannot draw the exact same ball twice; so the results of individual ball picks are mutually exclusive. So we use permutations without repetition to count the microstates for a given macrostate (the pair of ball colours). We can safely use combinations without repetition instead because all the microstates are paired.

Basically, the distinction is whether the diagonal is included or excluded.

$\begin{array}{c|cccccc}\text{dice}&1&2&3&4&5&6\\ \hline 1&2&3&4&5&6&7\\2&3&4&5&6&7&8\\3&4&5&6&7&8&9\\4&5&6&7&8&9&10\\5&6&7&8&9&10&11\\6&7&8&9&10&11&12\end{array}$ $\begin{array}{c|ccccccccc}\text{balls}&w&w&w&w&w&w&b&b&b\\\hline w&~&ww&ww&ww&ww&ww&wb&wb&wb\\ w&ww&~&ww&ww&ww&ww&wb&wb&wb\\ w&ww&ww&~&ww&ww&ww&wb&wb&wb\\ w&ww&ww&ww&~&ww&ww&wb&wb&wb\\ w&ww&ww&ww&ww&~&ww&wb&wb&wb\\ w&ww&ww&ww&ww&ww&~&wb&wb&wb\\b&bw&bw&bw&bw&bw&bw&~&bb&bb\\b&bw&bw&bw&bw&bw&bw&bb&~&bb\\b&bw&bw&bw&bw&bw&bw&bb&bb&~\end{array}$

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