How do I know if I should select simultaneously or separately? When using combinations to calculate how many ways an event occurs, how do I know whether to select objects simultaneously or separately?
For example, I am given the task of finding the probability of rolling a two-pair in poker dice. A two-pair is when, out of five rolled dice, two numbers appear exactly twice and one number appears once.
When determining the number of ways to select the two numbers that appear twice, why should I perform ${6 \choose 2}$ and not ${6 \choose 1}{5 \choose 1}$?
But when it comes time to choose the locations for these two numbers to appear, why must I calculate separately? Why do I perform ${5 \choose 2}{3 \choose 2}$ and not ${5 \choose 4}$?
Any help would be appreciated.
Edit: Is the choice of locations essentially a multinomial? So it would be ${5 \choose 2 2 1}$?
Edit 2: Should I be thinking along the lines of combining like terms?
 A: You can select the numbers individually, what matters is distinction between selections. Take your example $6 \choose 2$ vs ${6 \choose 1}{5 \choose 1}$ The first chooses 2 elements from 6, without order. The second chooses 1 element from 6, and then chooses a second element from the remaining 5 to pair with it. However, this establishes a distinction between the two, one was chosen first, and the other second. To remove this, we'd need to divide by the number of permutations of out resulting pair, $2!$, in which case we arrive at the same value as $6 \choose 2$.
The same applies to your ${5 \choose 2}{3 \choose 2}$ case. Here, we care about distinction- 2 of these elements belong to the first dice pair, 2 to the second. We don't care about distinction within a dice pair, hence why its not ${5 \choose 1}{4 \choose 1}{3 \choose 1}{2 \choose 1}$.
A: For your first question, $\binom{6}{2}$ implies choosing a group of two numbers which you want to appear twice each.
Say you perform $\binom{6}{1}$$\binom{5}{1}$. This will involve some overcounting: You will have
 CASE 1. You start out by choosing a number “a” from the 6 possible choices. Now there are 5 left. You choose a number “b” from them. You make combinations on the dice after choosing a third number “c”.
 CASE 2. You start out by choosing $“b”$ from the 6 possible choices. Now there are 5 left. You choose $”a”$from them. You make combinations on the dice after choosing a third number “c”.
But case 1 and 2, even though they are being counted separately in $\binom{6}{1}$$\binom{5}{1}$, represent the same set of combinations. Thus  you should have $\frac{1}{2}\binom{6}{1}\binom{5}{1}$, which is equal to $\binom{6}{2}$.
For the second question, if you perform $\binom{5}{4}$, you then must also take into account the various permutations of the two numbers (repeated twice each) on the four chosen dice. This implies the final answer is $\binom{5}{4}$.$\frac{4!}{2!2!}$ (considering repetitions) which is equal to $\binom{5}{2}$$\binom{3}{2}$.
 We are counting separately in the second part because permutations matter in the second part, i.e. which number goes on which die. In the first part, which number is selected before is irrelevant, so we count them together.
A: 
When determining the number of ways to select the two numbers that appear
twice, why should I perform $\binom{6}{2}$ and not $\binom{6}{1}\binom{5}{1}$?

Note that $\binom{6}{2} = 15$ and $\binom{6}{1}\binom{5}{1} = 30$. The definition of $\binom{6}{2}$ is "6 choose 2". Let's say you have five dice: $D_1, D_2, D_3, D_4$ and $D_5$.
The number of ways of selecting the two numbers that appear twice are
$$\frac{6*5}{2} = \binom{6}{2} = 15$$
Let's say I choose 5 first to be the first number to appear twice and then choose 6 to be the second. I can also choose 6 to be the first number and then choose 5 to be the second. Thus we divide $6*5$ by 2 to reduce the number of ways of choosing by half. If I choose $\binom{6}{1}\binom{5}{1}$ ($= 30$) it would account to both the cases of choosing 5 and 6 as mentioned above.

But when it comes time to choose the locations for these two numbers to appear, why must I calculate separately? Why do I perform $\binom{5}{2}\binom{3}{2}$ and not $\binom{5}{4}$?

In this case we have to choose 2 locations irrespective of order. It doesn't matter whether I choose $D_1$ first or $D_2$ first out of $D_1$ or $D_2$.
Out of any 4 dice we choose we can have any two die to share a number and the other two to share the other number. Thus if we choose 4 die out of 5 to share the two numbers (which corresponds to $\binom{5}{4}$) there is only 5 combinations to choose from but each of those combinations have $\binom{4}{2}$ ways of having 2 dice share 2 numbers each.
Note that the number of ways of choosing the locations for the two pair is
\begin{align}
\binom{5}{2} \binom{3}{2} &= \binom{5}{4} \binom{4}{2}
\end{align}
where $\binom{5}{4}$ ($=5$) is the number of ways of choosing 4 dice out of 5 and $\binom{4}{2}$ ($=6$) is the number of ways of choosing 2 of the 4 dice to share the same number.
You can check this article that helps understand poker dice which was also my source for understanding the two pair.
