PC: 6 people

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advita: Senior | Next Rank: 100 Posts; Posts: 62; Joined: Mon Sep 13, 2010 8:06 pm; Thanked: 2 times

by advita » Sun Jan 16, 2011 7:41 am

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.

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Source: Beat The GMAT — Problem Solving | Sun Jan 16, 2011 7:41 am

anshumishra: Legendary Member; Posts: 543; Joined: Tue Jun 15, 2010 7:01 pmFollowed by: 3 members; Thanked: 147 times

by anshumishra » Sun Jan 16, 2011 8:16 am

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.

Last edited by anshumishra on Sun Jan 16, 2011 8:27 am, edited 2 times in total.

Thanks
Anshu

(Every mistake is a lesson learned )

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RACHVIK: Senior | Next Rank: 100 Posts; Posts: 95; Joined: Wed Sep 22, 2010 8:26 amFollowed by: 1 member; Thanked: 1 times

by RACHVIK » Sun Jan 16, 2011 8:22 am

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

Rachvik

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anshumishra: Legendary Member; Posts: 543; Joined: Tue Jun 15, 2010 7:01 pmFollowed by: 3 members; Thanked: 147 times

by anshumishra » Sun Jan 16, 2011 8:43 am

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.

Thanks
Anshu

(Every mistake is a lesson learned )

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Ramit88: Senior | Next Rank: 100 Posts; Posts: 67; Joined: Fri Nov 26, 2010 8:25 am; Thanked: 3 times

by Ramit88 » Sun Jan 16, 2011 10:17 am

nice question.. very gmat like..

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pesfunk: Master | Next Rank: 500 Posts; Posts: 286; Joined: Tue Sep 21, 2010 5:36 pmFollowed by: 5 members; Location: Kolkata, India; Thanked: 11 times

by pesfunk » Thu Jan 27, 2011 6:33 am

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.

Like this post

aleph777: Master | Next Rank: 500 Posts; Posts: 131; Joined: Fri Jun 18, 2010 10:19 am; Location: New York, NY; Thanked: 10 times

by aleph777 » Thu Jan 27, 2011 7:12 am

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!

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Thu Jan 27, 2011 3:53 pm

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

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