anshumishra wrote:Lets try with a simpler example to understand it : 4 people 2 teams of 2 persons to be selected :
A B C D -> are the 4 person
Team 1------ Team 2
AB ------------ CD (after selecting AB we are left with CD)
AC -------------BD (after selecting AC we are left with BD)
AD -------------BC (after selecting AD we are left with BC)
Now, assuming team 1 and team 2 don't differ in anyway, we can't interchange the teams (otherwise it will have count the same combination twice)
So, it is 4C2/2! , in this case.
Such a helpful simplification, Anshumishra! I'm starting to get the hang of these combinatorics, but I just spent the last 10 minutes trying to apply this process to a hypothetical, modified problem without success.
How would you deal with something like this (it's not an official question, but rather just a way to better understand the concept above):
There are 9 people in a room, and you need to put them into three groups of three. How many ways can you combine?
I tried 9!/3!6! to get a first group, and then (6!/3!3!)/2! to get the other two groups and multiplied the two products together for a total of 840 possible combinations of groups of 3. Would that be a correct approach?
Thanks!
