“Equal Division of objects into groups” formula

Question

My book says the number of ways to distribute to 2n objects equally among two groups where order is considered is $\frac{2n!}{(n!)^2}$ but I have a doubt

Let's take an example

No. Of objects= 4 $\{A, B, C, D\}$

Number of ways to distribute these four object into groups which contain two objects each G1(_ ) G2( _) Now let's make all possible groups

 **G1**             **G2**
    AB.  |   AC, AD, BC, BD, CD
    AC.  |   AB, AD, BC, BD, CD
    AD.  |   AB, AC, BC, BD, CD
    BC.  |   AB, AC, AD, BD, CD
    BD.  |   AB, AC, AD, BC, CD
    CD.  |   AB, AC, AD, BC, BD

So, the number of ways to distribute these objects into two groups each having two objects when order of groups is considered is 6×5= 30 ways

As, $(AB, AC) (AB, AD) (AB, BC) (AB, BD) (AB, CD)$ 5 groups from each row above

**But according to the formula it is ** 4!/(2!×2!)= 6 ways

Please explain how this formula is true.

Answer 1

There are $(2n)!$ ways to line up the objects, because each way to do it is a permutation of these objects. The $n$ first objects in the row form group 1 and the $n$ last objects form group 2. There are $n! n!$ copies of the same distribution in two groups in this because each group can be permutated. Hence the result \begin{equation} \frac{(2n)!}{(n!)^2} \end{equation} Note that if you don't differentiate the groups (by their number) you need to divide this result by 2.