Whats the intuitive difference between ${n \choose 2}$ and taking ${n \choose 1} {{n-1} \choose 1}$? Example: I am trying to find the number of combinations of full house in poker. In that case I take ${13 \choose 1}{4 \choose 3} {12 \choose 1}  {4 \choose 2}$.
In a different scenario suppose I want to find the number of combinations of two pairs. There I take ${13 \choose 2}  {4\choose 2}  {4 \choose 2}$.
Why shouldn't I take ${13 \choose 1}$ and then ${12 \choose 1}$ like in the previous case?
 A: In a full house, there’s an asymmetry between the two values making up the poker hand - one occurs three times, and one occurs twice. In two pair, both values appear the same number of times. This means that if I tell you “I have two pair made of kings and tens,” you know what my hand is, but if I say “I have a full house made of kings and tens” you can’t tell whether I have three kings and two tens or three tens and two kings.
Mathematically, $\binom{13}{2}$ represents “pick two different values, and order doesn’t matter,” which is great for a two pair where the two different values don’t have an associated order. On the other hand, $\binom{13}{1}\binom{12}{1}$ means “pick some value as the first value, then pick some other value as the second value” means that order does matter, since one of them is specifically picked as the three-of-a-kind and one is picked as the two-of-a-kind.
Hope this helps!
A: The number $\binom{n}{2}$ is the number of ways to pick a $2$-element subset $\{a,b\}$ from a set $X$ of size $n$, while $\binom{n}{1}\binom{n-1}{1}$ is the number of ways of first picking an element $a$ from $X$ and then choosing an element $b$ from the set $X\setminus \{a\}$. In both cases, you end up with a $2$-element subset of $X$ but when we chose the 2-elements the first way, the order didn't matter. However, in the second way of choosing the elements, the order did matter. For instance, with the first method, you may have chosen the set $\{a,b\}$ while with the second method you may have picked $a$ and then $b$, or $b$ and then $a$. This is why there are in fact twice as many ways to choose $2$-elements in an ordered way versus in an unordered way.
