Same probability question, understood the sequential approach but not the counting approach I understand a similar question has been asked here, but it is dealing with same coloured balls; also it didn't seem to go specifically into the logic of counting method.
Probability that the withdrawn balls are the same color

This question comes from Bertsekas' Introduction to probability, 2nd ed.

An urn contains $m$ red and $n$ white balls. We draw two balls randomly and simultaneously. Calculate the probability that the selected balls are of different color, by using
two approaches: a counting approach based on the discrete uniform law, and a sequential approach based on the multiplication rule.


I actually managed to get the correct answer with both approaches; what I am struggling with is while I understand the logic of sequential approach, I can't say the same with the counting approach.
Sequential:
As per the sequential diagram, it's $\frac{m}{m+n}\times\frac{n}{m+n-1}+\frac{n}{m+n}\times\frac{m}{m+n-1}$ in virtue of the total probability theorem, which comes to $\frac{2mn}{m+n-1}(m+n)$.

Counting:
I managed to guess it, namely $\frac{\binom{m}{1}\binom{n}{1}}{\binom{m+n}{2}}$, which also comes to $\frac{2mn}{(m+n-1}(m+n)$.
My logic: $\binom{m}{1}$ refer to the number of ways to pick 1 ball out of $m$ many reds (likewise for $\binom{n}{1}$).
My understanding is that $\binom{n}{k}$ counts the number of combinations when one picks $k$ elements out of $n$ elements. e.g. If $m=2$, then we have 1) $?R$ and 2) $R?$ as all combinations possible, giving a total of $2$. ($?$ could be another $R$ or $W$)
$\binom{m+n}{2}$ refers to the total number of possible combinations of balls.
What I don't understand is that this doesn't seem to take order into account, i.e. shouldn't permutation be involved here? In particular, $\binom{m}{1}\times\binom{n}{1}$ feels as if we are always picking the red ball first, and then picking whites - but surely we can also pick the white first and then red, since $WR$ counts just as $RW$ does?
 A: I would explain it like this. Here are three ways to think about it: The approaches can all be said to use 'counting' if you phrase them like that but to my mind the important point is that they are characterized by what orders are being taken into account when you do the counting.
1. The actual real truth: Balls are different and order matters. This is like imaging that beforehand you actually number them $1,...,n, n+1,...,n+ m$. So just like in real life, the balls are all actually different objects and ball 1 followed by ball $n+3$ is different from pulling ball $n+7$ followed by ball $2$, even though both of those situations result in a differently coloured pair of balls having been picked.
So, thinking along these lines, the denominator is number of ways of picking any two balls from the $m+n$ distinct balls that you have, which is
$$
(m+n)(m+n-1)
$$
The numerator is the number of ways of picking one red ball and one white ball. If I pick red first, there are $m$ options for the red ball and for each choice of red ball there are $n$ possibilities for the white ball. I must add to this the number of ways of picking white first and then red, which is $nm$ so there are $2mn$ ways of picking differently coloured balls.
$$
\text{Answer} = \frac{2mn}{(m+n)(m+n-1)}.
$$
2. 'Sequential' Approach:
I'll skim this a bit since you got it right. Using the law of total probability:
\begin{eqnarray}
& \mathbb{P}\Bigl(\text{Select different colour balls}\Bigr) \\
&= \mathbb{P}\Bigl(\text{Select different colour balls} \cap \text{White first}\Bigr) + \mathbb{P}\Bigl(\text{Select different colour balls} \cap \text{Red first}\Bigr)
\end{eqnarray}
I could stop here and compute each of those two terms by counting. I'd get
$$
\frac{mn}{(m+n)(m+n-1)} + \frac{mn}{(m+n)(m+n-1)},
$$
because as we saw above, the number of ways of picking differently coloured ball in which you pick red first is $mn$ and same if you pick white first. Or we can continue using conditional probability, to really see the 'sequential' way of thinking:
\begin{eqnarray}
&= \mathbb{P}\Bigl(\text{Select different colour balls} | \text{White first}\Bigr)\mathbb{P}(\text{White first}) + \mathbb{P}\Bigl(\text{Select different colour balls} | \text{Red first}\Bigr)\mathbb{P}(\text{Red first}) \\
&= \mathbb{P}\Bigl(\text{Select different colour balls} | \text{White first}\Bigr)\frac{n}{m+n} + \mathbb{P}\Bigl(\text{Select different colour balls} | \text{Red first}\Bigr)\frac{m}{m+n} \\
&= \mathbb{P}\Bigl(\text{Second ball is Red} | \text{White first}\Bigr)\frac{n}{m+n} + \mathbb{P}\Bigl(\text{Second ball is White} | \text{Red first}\Bigr)\frac{m}{m+n} \\ 
&= \frac{m}{m+n-1}\frac{n}{m+n} + \frac{n}{m+n-1}\frac{m}{m+n} \\ 
&= \frac{2mn}{(m+n)(m+n-1)}
\end{eqnarray}
3. 'Counting Approach': Order does not matter. Here we are not even imagining that we reach our hand in and pick one ball before the next. We are only asking for the demoninator: How many sets of 2 balls are there from the collection $m+n$ balls?
$$
\binom{m+n}{2} = \frac{(m+n)!}{(m+n-2)!2!} = \frac{1}{2}(m+n)(m+n-1)
$$
For the numerator, how many sets of two differently coloured balls are there from the collection of $m+n$ balls? To give a different spin on it, notice that this has to be the size of all the subsets  of 2 balls minus the subsets that are both red  and minus the subsets that are both white, so we have
\begin{eqnarray}
& \frac{1}{2}(m+n)(m+n-1) - \binom{m}{2} - \binom{n}{2} \\
&= \frac{1}{2}\Bigl((m+n)(m+n-1) - m(m-1) - n(n-1)\Bigr) \\
& = mn
\end{eqnarray}
Giving
$$
\text{Answer} = \frac{mn}{\frac{1}{2}(m+n)(m+n-1)}
$$
In these problems, really you choose a different probability space depending on which "approach" you are taking. So for number 3, it's as if you choose the probability space to consist of 2-sets from the collection of $m+n$ balls, and it is impossible to distinguish the ordering within a given 2-set. Whereas in number 1. and 2., the probability space consists of tuples like (1, $m+2$), meaning you picked ball 1 first (red) and then ball $m+2$ (red).
