How do we know when order matters? The following is from Ross "Introduction to probability and statistics for engineers and scientists" Example 3.5.4
We choose 6 pairs of room mates from a team of 12 basketball players.
It follows from the generalized basic principle of counting that there are
$$\left(\begin{array}{c}12 \\ 2\end{array}\right)\left(\begin{array}{c}10 \\ 2\end{array}\right)\left(\begin{array}{l}8 \\ 2\end{array}\right)\left(\begin{array}{l}6 \\ 2\end{array}\right)\left(\begin{array}{l}4 \\ 2\end{array}\right)\left(\begin{array}{l}2 \\ 2\end{array}\right)=\frac{12 !}{(2 !)^{6}}$$
ways of dividing the players into a first pair, a second pair and so on.
From my question here I understand that we then divide by $6!$ to get the number of unordered pairs.
However it is not intuitive to me that order matters when we divide into "a first pair, a second pair and so on"
I do get that we are creating six pairs one at a time.
Looking at my probability cheat sheet

It seems to me that we are using the formula from bottom right corner where order does not matter.
Where could I be going wrong?

$12 \choose 2$ is the number of ways of choosing 2 out of 12 (without replacement) when order doesn't matter.  That is, $12 \choose 2$ accounts for order not mattering within a pair.
In the process of 'choose a first pair, choose a second pair,...' order does matter: we distinguish the first pair, chosen out of 12 possibilities, from the second pair, chosen out of the remaining 10 possibilities, and so on. So, $${12 \choose 2}{ 10 \choose 2}{ 8 \choose 2}{ 6 \choose 2}{ 4 \choose 2}{ 2 \choose 2}$$
is the number of ways of choosing a first (unordered) pair, then a second (unordered) pair, and so on. The pairs are different; Chris and Sandy in the last pair know they were chosen last and they feel bad about it.
The question has no ordering between pairs. If Chris and Sandy were chosen first, they would be happier but Professor Ross wouldn't care. So we need to divide by the $6!$ ways of ordering between pairs.
You could do the same calculation differently. For example, you could start with a purely ordered format: stand the players in line and pair off 1 and 2, 3 and 4, and so on.  This gives $12!$ completely ordered possibilities, and you need to divide by $2^6$ for the 2 orderings within each of six pairs, and $6!$ for the orderings of  the pairs.  I think this is simpler.
I don't see a simple way to do the whole calculation as unordered from the start, but maybe someone else will suggest one
