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), ${n \choose 2} \cdot {n-2 \choose 2} \cdot \dfrac{1}{2}\neq {n \choose 4}$?
 A: With $\binom n2\binom{n-2}2/2$ you are choosing four elements, but then you are splitting that selection into two unordered groups of two. Hence it is doing more than merely selecting four elements ($\binom n4$), and produces a larger number.
A: Consider $n=6$, and assume that set $S = \{1,2,3,4,5,6\}.$
Consider the specific collection of 4 items represented by the subset $T = \{1,2,3,4\}.$
There are 3 distinct ways of partitioning set $T$ into two pairs: $\{(1,2),(3,4)\}$, $\{(1,3),(2,4)\}$, and $\{(1,4),(2,3)\}$.
This means that since there are $\binom{6}{4} = 15$ ways of forming a collection of four items (similar to subset $T$) from set $S$, there are $(3 \times 15) = 45$ ways of partitioning set $S$ into two distinct pairs, with two items left over.
Superficially, you would expect that the # of ways of forming the two distinct pairs from set $S$ would be 
$\displaystyle \binom{6}{4} \times \binom{4}{2}$, (which is wrong) 
rather than $\displaystyle \binom{6}{4} \times \binom{4}{2} \times \frac{1}{2}$, (which is right).  The last factor, $\frac{1}{2}$, which is an overcounting adjustment scalar is needed, because otherwise each of the partitions into two pairs would be double counted.
As an illustration of this, note that the partitioning of set $T$ above, into $2$ pairs could only be done in $3$ ways, even though $\binom{4}{2} = 6.$
In general, there will be $\binom{n}{4}$ ways of forming collections of $4$ items from a set of $n$ items, and there will be 
$\displaystyle \binom{n}{4} \times \binom{4}{2} \times \frac{1}{2}$ ways of forming two distinct pairs of two items each, from a set of $n$ items.
A: 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:
$$\binom{n}{2}\binom{n-2}{2}=\binom{n}{4}\binom{4}{2}$$
