Breaking the solution to the problem of dealing one pair just to see what happens The answer is $\binom {13}1 \binom42 \binom{12}3 \binom 41^3$
I want to break the last term and see what happens. [Struggling with the concept so trying to work with it as much as possible].
$\binom 41^3$ means that $\heartsuit \diamondsuit \spadesuit$ is different from $\diamondsuit \heartsuit \spadesuit$ and $\heartsuit \spadesuit \diamondsuit $ 
$\binom 42 \binom 41$ means that $\heartsuit \diamondsuit \spadesuit$ is is the same as $\diamondsuit \heartsuit \spadesuit$
$\binom 43$ means that $\heartsuit \diamondsuit \spadesuit$,  $\diamondsuit \heartsuit \spadesuit$, $\heartsuit \spadesuit \diamondsuit $ are all the same.
Basically, in the first case all three suits are ordered, in the second case first two suitss are unordered but the third one is and in the third case none of the suits are ordered.
Does that make sense?
Clarification:
A one pair consists of five cards where two are of the same kind and the other three are of different kinds. How many such hands are possible? The correct answer is $\binom {13}1 \binom42 \binom{12}3 \binom 41^3$. What I am doing is choosing the suits in different ways to see how that affects the correct answer. Want to see if my reasoning holds up.
 A: Before discussing the last factor, note that the second last factor, $\binom{12}3$ counts the ways to select three from twelve values without repetition.
Now, $\binom{4}{1}^3$ counts ways to pick three suits with repetition.   We need to match these suits to three distinct numbers, so the fact that this counts distinct order arrangements is okay. 
$$\{\heartsuit\heartsuit\heartsuit,\heartsuit\heartsuit\spadesuit,\heartsuit\heartsuit\clubsuit,\heartsuit\heartsuit\diamondsuit, \heartsuit\spadesuit\heartsuit,\ldots,\diamondsuit\diamondsuit\diamondsuit\}$$  

If we used $\binom{4}3$ we would be selecting three from four suits, without repetition.
$$\{\heartsuit\spadesuit\clubsuit, \heartsuit\spadesuit\diamondsuit, \heartsuit\clubsuit\diamondsuit,\spadesuit\clubsuit\diamondsuit\}$$
To properly match these with the three distinct values, we should multiply by $3!$ ways to arrange three distinct suits.   $\binom {12}3\binom 4 3 3!$ counts the ways to select three cards each of distinct value and distinct suit, selected from twelve values and four suits.

However, $\binom 42\binom 41$ attempts to count ways to select two distinct suit (without repetition) then a third suit which may be a repetition, which miscounts.   Instead we should count: $\binom 4 3 3!+\binom 42\binom 21 3$ ways to select three distinct suits or a pair and singleton of suits, to be matched to the three distinct values.
