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

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In how many different ways can a group of 8 people be divide [#permalink] New post 24 Oct 2009, 03:20
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A
B
C
D
E

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Question Stats:

30% (01:00) correct 70% (00:58) wrong based on 211 sessions
In how many different ways can a group of 8 people be divided into 4 teams of 2 people each?

A, 90
B. 105
C. 168
D. 420
E. 2520
[Reveal] Spoiler: OA

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Re: combination (groups and that stuff...) [#permalink] New post 24 Oct 2009, 16:14
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
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Re: combination (groups and that stuff...) [#permalink] New post 24 Oct 2009, 16:20
Another approach:

Number of ways 8 people can be arranged = 8!
Number of different teams possible = 8!/2^4 = 2520
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Re: combination (groups and that stuff...) [#permalink] New post 24 Oct 2009, 17:50
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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.
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Re: combination (groups and that stuff...) [#permalink] New post 24 Oct 2009, 17:56
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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.

Answer: A.

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

Answer: B.
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Re: combination (groups and that stuff...) [#permalink] New post 24 Oct 2009, 17:58
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Re: combination (groups and that stuff...) [#permalink] New post 24 Oct 2009, 18:53
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....
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Re: combination (groups and that stuff...) [#permalink] New post 24 Oct 2009, 19:46
OMG, I completely missed that we also need to divide by 4!. Thanks for explaining.
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Re: combination (groups and that stuff...) [#permalink] New post 25 Oct 2009, 04:08
Thanks for correction.....I missed it completely..
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Re: Permutations/ Combinations [#permalink] New post 03 Dec 2009, 04:49
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
AD BC

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?
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Re: combination (groups and that stuff...) [#permalink] New post 17 Oct 2010, 18:29
B
8!/(2!2!2!2!)*4!

4! because the order does not matter
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Re: In how many different ways can a group of 8 people be divide [#permalink] New post 29 Oct 2013, 02:13
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Re: In how many different ways can a group of 8 people be divide [#permalink] New post 16 Nov 2013, 22:45
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The answer is B - 105 but the OA shows A. Please correct this.
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Re: In how many different ways can a group of 8 people be divide [#permalink] New post 17 Nov 2013, 03:25
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Re: combination (groups and that stuff...) [#permalink] New post 18 Nov 2013, 17:36
Two questions are asked;
First the no of ways in which 2 can e selected fro 8 peoples;
= 8C2 * 6C2 * 4C2* 2C2
=(8!/6!*2!)*(6!/4!*2!)*(4!/2!*2!)*(2!/2!*0!)
=2520

Second the no of ways in which 2 can be arranged among 4 groups ; since the arrangement will vary hence simple factorial of will be the choice which is 4!

So finally you need to divide the former by later to get the answer
=2520/4!
=105
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Re: combination (groups and that stuff...) [#permalink] New post 30 Dec 2013, 05:03
IanStewart wrote:
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.


This is great to clarify Ian

So basically we have three levels among permutations and combinations we have:

Permutation where order of elements does matter for instance {A,B,C} is different from {B,C,A} and so on
Then we have combinations in which order does not matter in these cases {A,B,C} = {C,B,A}, but if we have two groups then say group 1 is {A,B,C} and group 2 is {D,E,F} then {1,2} is different from {2,1}, as you explained in your example with the tennis couples going to different tournaments. So the order of the sets themselves ARE important.

Now if we DON'T care about the order of the arrangements of the SETS (not the elements within the sets), then we get to our third level in which we need to divide by n! (n being the number of GROUPS) so that we are stating no preference over order of sets.

I hope I'm on the right track

Let me know if this sounds OK

Cheers!
J :)
Re: combination (groups and that stuff...)   [#permalink] 30 Dec 2013, 05:03
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