Counting question - binomial coefficients Why is there a difference in choosing $2$ out of $n$ elements and after that $2$ out of the $n-2$ remaining elements compared to choosing $4$ elements. So why is, intuitively (not algebraically), ${n \choose 2} \cdot {n-2 \choose 2} \cdot \dfrac{1}{2}\neq {n \choose 4}$?

The first method selects two groups of two and not one of four. So $\{a,b,c,d\}$ could be $\{a,b\}\{c,d\}$, or $\{a,c\}\{b,d\}$, and so on.
In fact, we have:
$$\binom{n}{2}\binom{n-2}{2}=\binom{n}{4}\binom{4}{2}$$
