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?
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.