Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 18 May 2013, 09:20

In how many different ways can a group of 8 people be

Author Message
TAGS:
SVP
Joined: 16 Jul 2009
Posts: 1635
Schools: CBS
WE 1: 4 years (Consulting)
Followers: 25

Kudos [?]: 118 [0], given: 2

combination (groups and that stuff...) [#permalink]  24 Oct 2009, 04:20
00:00

Question Stats:

75% (01:34) correct 25% (00:50) wrong based on 1 sessions
In how many different ways can a group of 8 people be divided into 4 teams of 2 people each?

90
105
168
420
2520
_________________

The sky is the limit
800 is the limit

Find out what's new at GMAT Club - latest features and updates

Intern
Joined: 16 Oct 2009
Posts: 11
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: combination (groups and that stuff...) [#permalink]  24 Oct 2009, 17:14
4 teams can be chosen in following ways,
= 8C2 * 6C2 * 4C2* 2C2
=(8!/6!*2!)*(6!/4!*2!)*(4!/2!*2!)*(2!/2!*0!)

Solving it further you will get
= (28) * (15) * (6) * (1)
= 2520

E is ans
Senior Manager
Joined: 18 Aug 2009
Posts: 340
Followers: 5

Kudos [?]: 120 [0], given: 13

Re: combination (groups and that stuff...) [#permalink]  24 Oct 2009, 17:20
Another approach:

Number of ways 8 people can be arranged = 8!
Number of different teams possible = 8!/2^4 = 2520
GMAT Instructor
Joined: 24 Jun 2008
Posts: 973
Location: Toronto
Followers: 167

Kudos [?]: 443 [2] , given: 3

Re: combination (groups and that stuff...) [#permalink]  24 Oct 2009, 18:50
2
KUDOS
The answer to the question is 105, not 2520. I posted a solution on another forum, which I'll paste here:

________

Think of this question:

A group of eight tennis players will be divided into four teams of two. One team will play in the Olympics, one in Wimbledon, one in the Davis Cup and one in the US Open. In how many different ways can the teams be selected?

Here, the order of the teams themselves clearly matters. If we choose {A,B} to go to the Olympics, and {C,D} to go to Wimbledon, that's clearly different from sending {C,D} to the Olympics and {A,B} to Wimbledon. The answer to this question is exactly the answer you give above:

-you have 8C2 choices for the Olympics team;
-you have 6C2 choices for the Wimbledon team;
-you have 4C2 choices for the Davis Cup team;
-you have 2C2 (one) choice for the US Open team.

Multiply these to get the answer: 8C2*6C2*4C2*2C2 = (8*7/2)(6*5/2)(4*3/2)(2*1/2) = 2520.

Note that the question I've just asked above is different from the question in the original post. In this question:

A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

the order of the teams does not matter. If we choose, say, these teams:

{A,B}, {C,D}, {E,F}, {G,H}

that's exactly the same set of teams as these:

{C,D}, {A,B}, {G,H}, {E,F}

Because the order of the teams themselves does not matter, we must divide by 4! = 24, the number of different orders we can put the four teams in, because all 24 different orders are in fact the same set of teams. So the answer is 2520/4! = 105.
_________________

Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details.

Private GMAT Tutor based in Toronto

GMAT Club team member
Joined: 02 Sep 2009
Posts: 11506
Followers: 1791

Kudos [?]: 9518 [1] , given: 826

Re: combination (groups and that stuff...) [#permalink]  24 Oct 2009, 18:56
1
KUDOS
I think there is a little problem with the solutions above:

Does the order matters? Think not.
(1,2)(3,4)(5,6)(7,8) should be the same 4 teams as (5,6)(7,8)(1,2)(3,4), as we don't have team #1, team #2...

So 2520 should be divided by 4!=105.

I know there is a formula to determine:

A. the number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is not important.
B. the number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is important.

Basically hgp2k used here the second one, but these formulas are not needed for GMAT and there is an easier way to solve this problem, well at least I solve this way and find it easier:

For the first person we can pick a pair in 7 ways;
For the second one in 5 ways (as two are already chosen);
For the third one in 3 ways (as 4 people are already chosen);
For the fourth one there is only one pair left.

So we have 7*5*3*1=105

_________________
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11506
Followers: 1791

Kudos [?]: 9518 [0], given: 826

Re: combination (groups and that stuff...) [#permalink]  24 Oct 2009, 18:58
Sorry, didn't notice that Ian posted the right solution just 6 min before I did.
_________________
Senior Manager
Affiliations: PMP
Joined: 13 Oct 2009
Posts: 319
Followers: 2

Kudos [?]: 76 [0], given: 37

Re: combination (groups and that stuff...) [#permalink]  24 Oct 2009, 19:53
I knew the answer was 105 and that we have to divide by 2520 by 4!, but not sure why to divide by 4!.

Thanks Ian for explaining that....
_________________

Thanks, Sri
-------------------------------
keep uppp...ing the tempo...

Press +1 Kudos, if you think my post gave u a tiny tip

Senior Manager
Joined: 18 Aug 2009
Posts: 340
Followers: 5

Kudos [?]: 120 [0], given: 13

Re: combination (groups and that stuff...) [#permalink]  24 Oct 2009, 20:46
OMG, I completely missed that we also need to divide by 4!. Thanks for explaining.
Intern
Joined: 16 Oct 2009
Posts: 11
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: combination (groups and that stuff...) [#permalink]  25 Oct 2009, 05:08
Thanks for correction.....I missed it completely..
Intern
Joined: 14 Sep 2009
Posts: 26
Location: New York, NY
Followers: 0

Kudos [?]: 7 [0], given: 5

Re: Permutations/ Combinations [#permalink]  03 Dec 2009, 05:49
Economist wrote:
I cannot comprehend this concept
Lets say there are 4 people, ABCD to be arranged in two groups of two. There are 6 ways of doing this:
AB, AC, AD, BC, BD and CD.
If we use the above formula we get 4C2*2C2 / 2! = 6/2 = 3 !!!
What is the difference ??!!??

I was struggling with the question as well and wanted to use similar logic. Here's my thought - the question asks In how many different ways can a group of 8 people be divided into 4 teams of 2 people each?

Which, IMO, is slightly different from "How many teams of two can we create from a group of 8 people?" The first question assumes that one string of teams is one unit.

Using the example above,

AB CD
AC BD

Gives us three units of teams (6/2!) = 3. I think it's just a different way of looking at it - we have 6 teams but only three possible SETS of teams - therefore, we additionally have to divide by the factor of the number of teams.

Thoughts on this?
Manager
Joined: 25 Jul 2010
Posts: 177
WE 1: 4 years Software Product Development
WE 2: 3 years ERP Consulting
Followers: 4

Kudos [?]: 18 [0], given: 15

Re: combination (groups and that stuff...) [#permalink]  17 Oct 2010, 19:29
B
8!/(2!2!2!2!)*4!

4! because the order does not matter
_________________

Re: combination (groups and that stuff...)   [#permalink] 17 Oct 2010, 19:29
Similar topics Replies Last post
Similar
Topics:
How many different ways can a group of 8 be divided into 4 6 22 Jan 2006, 09:48
In how many different ways can a group of 8 be divided into 7 09 Nov 2007, 11:49
In how many different ways can a group of 8 be divided into 1 09 Mar 2008, 01:02
4 In how many different ways can a group of 8 people be 5 13 Aug 2010, 08:38
1 In how many different ways can a group of 8 people be 8 12 Jan 2012, 07:59
Display posts from previous: Sort by