TITLE: In how many ways can '$6$' things be distributed equally among $2$ groups?
QUESTION [6 upvotes]: In how many ways can '$6$' things be distributed 
  equally among $2$ groups ? 

I tried
$\dbinom{6}{3}\times \dbinom{3}{3}\times 3!\times 3!$
But I am not sure if it is correct .
I look for a short and simple way.
I have studied maths up to $12$th grade.

REPLY [7 votes]: It's $6\choose 3$ if you have groups $A$ and $B$.  You choose three items and give them to $A$; there are $6\choose 3$ ways to do this.  You then give the remaining to group $B$.  
The $3!s$ are called for if you stipulate the order in which the groups receive the items is important. Otherwise, not.  
$${6\choose 3} = 20.$$
If the two groups are indistinguishable and you are just interested in partitioning the items into two parts of size three, we have an overcount of a factor of 2.  In that case there are only $10$ ways.  You must be careful about exactly what you are counting.