PC: 6 people

advita wrote:Q: There are total six people & we want to form two committees of three people each.What are the no of ways of doing this?

pl explain the proceedings... thanks.

Select any 3 people for one of the team : 6C3 = 20 ways.
Now rest of the 3 people fall in the other team (automatically), so no need to select anything.
Assuming that the two teams don't differ in anyways , it should be 20/2! = 10.

We can form 6C3 or 20 groups of three people each. In order to divide them into 2 teams divide groups by 2!, so the answer should be 10 ways. Whats the OA.

thanx

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.

nice question.. very gmat like..

Could someone please explain why we divide the 6C3 by 2! . I am still confused with the last division part.

Thanks

anshumishra wrote:

advita wrote:Q: There are total six people & we want to form two committees of three people each.What are the no of ways of doing this?

pl explain the proceedings... thanks.

Select any 3 people for one of the team : 6C3 = 20 ways.
Now rest of the 3 people fall in the other team (automatically), so no need to select anything.
Assuming that the two teams don't differ in anyways , it should be 20/2! = 10.

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!

advita wrote:Q: There are total six people & we want to form two committees of three people each.What are the no of ways of doing this?

pl explain the proceedings... thanks.

The following thread shows different approaches to this sort of question:

https://www.beatthegmat.com/forming-teams-t73034.html