# “Equal Division of objects into groups” formula

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.

• Use mathjax to render the maths, you have a tuto here math.meta.stackexchange.com/questions/5020/… – Satyendra Jul 28 '20 at 15:02
• Why can't G2 contain AB in the first line? – JMP Jul 28 '20 at 15:03
• @JMP bcoz when we place A and B in 1st group they are already taken and we can't put them in 2nd group – Nikhil Pant Jul 28 '20 at 15:06
• Then you can't have AC in the second group, can you ? – Fabien Jul 28 '20 at 15:09
• @Fabien here i tried to show when I take A and B in 1st group then those in G2 are all other options of groups that can be in group 2 – Nikhil Pant Jul 28 '20 at 15:11

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 $$$$\frac{(2n)!}{(n!)^2}$$$$ Note that if you don't differentiate the groups (by their number) you need to divide this result by 2.