Solving a question about probability with ordered elements with combinations I'm trying to solve the question 6 of chapter 3 from A First Course in Probability, from Sheldon Ross:

Consider an urn containing 12 balls, of which 8 are white. A sample of
   size 4 is to be drawn with replacement (without replacement). What is
   the conditional probability (in each case) that the first and third
   balls drawn will be white given that the sample drawn contains exactly
   3 white balls?

I solved this question, but I didn't understand this solution from this book
Please see this picture, I tried to summarize my doubts below:

I want to understand his solution because I think it could be useful to solve other similar questions. So I have the following questions:


*

*Why does he uses combinations (the order doesn't matter), since the
balls are clearly ordered.

*Why in another moment he sees the balls as they are the same?

*Sometimes I see the authors solving questions with ordered elements with combinations, but I don't understand why, do you know a simple example to help me to understand this reasoning?

 A: From what I've understood, you know how to use the formula, but not why it is what it is.
Recall the definition of a probability: $\frac{\text{favourable outcomes}}{\text{possible outcomes}}$. Since the "favourable outcome" here is the sample containing three white and one non-white ball, we must find out in how many ways this is possible. If you have 8 in total and want to take out three, this can be done in ${8\choose 3}$ ways. Similarly, if you have 4 non-white objects and take out one of these, you can do it in ${4\choose 1}$ ways. Now, there are 12 objects in total and you take out four of these, which can be done in ${12\choose 4}$ ways.
Moving on, he does not see the balls as the same, but those two are the single 2 ways in which three white and one non-white are drawn (event $E$) and that the first and third ball are white (event $F$). The point is that the balls are "not the same", but the two outcomes {W,W,W,B} and {W,B,W,W} are the two only outcomes which satisfies $F\cap E$.
Please ask if anything is still unclear.
A: I am sorry if I am being extremely late. But I believe you are completely right in the sense that the balls should have been ordered based on their positions. P(E) is calculated correctly because in this case, order doesn't matter. We only need to make sure that out of the 4 balls chosen, 3 are white.
However, in the case of $$P(F \cap E)$$, order does matter, because we need to make sure the first chosen and the third chosen are indeed white. In this case, it should be $$\frac{{8 \choose 3} * 3!* {4 \choose 1} * 2}{{12 \choose 4} * 4!}$$
The reason why we did 8 choose 3 is because we need to choose 3 balls out of the 8 white balls. Similar to what you have, we only have two possible positions for this: {W, B, W, W} and {W, W, B, W}. In each case, we have 3! possible arrangement for the white balls. We also have 4 different options for the non-white balls, hence the 4 choose 1. The 2 at the end represents the 2 different positions.
Similarly, since the numerator is ordered, we need the total possible outcomes ordered as well, hence the * 4! at the end.
This will simplify down to $$\frac{112}{495}$$
The entirety of P (F | E) should have simplified down to simply 1/2 instead
