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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]

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24 Oct 2009, 03:20
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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
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Re: combination (groups and that stuff...)  [#permalink]

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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]

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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]

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24 Oct 2009, 16:20
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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]

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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.

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

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Re: combination (groups and that stuff...)  [#permalink]

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24 Oct 2009, 17:58
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Sorry, didn't notice that Ian posted the right solution just 6 min before I did.
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Re: In how many different ways  [#permalink]

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12 Jan 2012, 07:06
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14
manalq8 wrote:
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

$$\frac{C^2_8*C^2_6*C^2_4*C^2_2}{4!}=105$$, we are dividing by 4! (factorial of the # of teams) as the order of the teams does not matter. If 8 people are - 1, 2, 3, 4, 5, 6, 7, 8, then (1,2)(3,4)(5,6)(7,8) would be the same 4 teams as (5,6)(7,8)(1,2)(3,4), as we don't have team #1, team #2...

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 left.

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

You can check similar problems:
probability-88685.html?hilit=different%20items%20divided%20equally
probability-85993.html?highlight=divide+groups
combination-55369.html#p690842
sub-committee-86346.html?highlight=divide+groups

There is also direct formula for this:

1. 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 is $$\frac{(mn)!}{(n!)^m*m!}$$.

2. 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 is $$\frac{(mn)!}{(n!)^m}$$

Hope it helps.
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Re: In how many different ways can a group of 8 people be  [#permalink]

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01 Mar 2013, 06:07
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 left.

Hi Bunuel,

Can u tell y have u considered 7 ways above?

Also,in the previous method im not clear why we need to divide by 4!.
Is it like say for eg:how many ways can we arrange the word AEEB so we need to consider 4!/2! since EE is repeated?
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Re: In how many different ways can a group of 8 people be  [#permalink]

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01 Mar 2013, 20:09

Suppose you were choosing a person to pair with person 1. You could form the following pairs
(1,2),(1,3)(1,4),(1,5),(1,6),(1,7)(1,8)
That's a total of 7 choices possible. Hence, one needs to choose 7.

Hope this helps!

shreerajp99 wrote:
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 left.

Hi Bunuel,

Can u tell y have u considered 7 ways above?

Also,in the previous method im not clear why we need to divide by 4!.
Is it like say for eg:how many ways can we arrange the word AEEB so we need to consider 4!/2! since EE is repeated?
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Re: In how many different ways can a group of 8 people be  [#permalink]

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01 Mar 2013, 21:05
1
shreerajp99 wrote:
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 left.

Hi Bunuel,

Can u tell y have u considered 7 ways above?

Also,in the previous method im not clear why we need to divide by 4!.
Is it like say for eg:how many ways can we arrange the word AEEB so we need to consider 4!/2! since EE is repeated?

Let me try explain what I thought:
Person 1 to pair with someone - have 7 choices (out of 8)
Person 2 to pair with somone - have 5 choices (out of remaining 6 people, note the 2nd person is also included in remaining 6)
.......Likewise

Divide by 4!, you are close to correct, it is to avoid repeats of similar groups. Since order of the the chosen groups does not matter here (Person 1, Person 2) is same as (Person 2, Person 1) - that means as per the formula we have number of groups which includes these repeats, to negate those we divide by 4! to get a realistic number with no such repeats

I tried, hope it is clear.....

Thanks
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Re: In how many different ways can a group of 8 people be  [#permalink]

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03 Mar 2013, 05:31
Hmm Bunuel something I did quickly, and forgot the context to do this in (this doesn't happen as frequently) was to select by team? How would I approach the problem then. IE instad of 8C2, I started with 4C1 and started to proceed from there, with the hopes of multiplying by 2! to account for the different arrangements we could have within each team (but not by 4! to account for the order of these different teams)

Why would 4C1 not be appropriate in this case? Is it because those 4 teams aren't set beforehand?

I am messing this up conceptually and want to correct this mistake
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Re: In how many different ways can a group of 8 people be  [#permalink]

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03 Mar 2013, 23:02
2
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manimgoindowndown wrote:
Hmm Bunuel something I did quickly, and forgot the context to do this in (this doesn't happen as frequently) was to select by team? How would I approach the problem then. IE instad of 8C2, I started with 4C1 and started to proceed from there, with the hopes of multiplying by 2! to account for the different arrangements we could have within each team (but not by 4! to account for the order of these different teams)

Why would 4C1 not be appropriate in this case? Is it because those 4 teams aren't set beforehand?

I am messing this up conceptually and want to correct this mistake

Check out this post: http://www.veritasprep.com/blog/2011/11 ... ke-groups/

It discusses two different questions:
1. Distributing 12 different chocolates equally among 4 boys (similar to splitting 8 people in 4 distinct teams with 2 people each - your question)
2. Distributing 12 different chocolates equally in 4 stacks (similar to splitting 8 people in 4 teams of 2 people each - the original question)

See if grouping makes sense thereafter.
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Re: In how many different ways can a group of 8 people be  [#permalink]

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05 Mar 2013, 18:02
VeritasPrepKarishma wrote:
manimgoindowndown wrote:
Hmm Bunuel something I did quickly, and forgot the context to do this in (this doesn't happen as frequently) was to select by team? How would I approach the problem then. IE instad of 8C2, I started with 4C1 and started to proceed from there, with the hopes of multiplying by 2! to account for the different arrangements we could have within each team (but not by 4! to account for the order of these different teams)

Why would 4C1 not be appropriate in this case? Is it because those 4 teams aren't set beforehand?

I am messing this up conceptually and want to correct this mistake

Check out this post: http://www.veritasprep.com/blog/2011/11 ... ke-groups/

It discusses two different questions:
1. Distributing 12 different chocolates equally among 4 boys (similar to splitting 8 people in 4 distinct teams with 2 people each - your question)
2. Distributing 12 different chocolates equally in 4 stacks (similar to splitting 8 people in 4 teams of 2 people each - the original question)

See if grouping makes sense thereafter.

That example and wording was extremely confusing and frustrating. I am still trying to see what language prompted the difference in permutation vs combination.

Maybe my brain has been fried this week (lots of pracatice, one full length CAT), but the big thing in this problem is you DO NOT account for order TWICE
the 4! is for the team
and the 2! is within every single combination for each pair
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Re: In how many different ways can a group of 8 people be  [#permalink]

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05 Mar 2013, 19:09
manimgoindowndown wrote:
That example and wording was extremely confusing and frustrating. I am still trying to see what language prompted the difference in permutation vs combination.

Maybe my brain has been fried this week (lots of pracatice, one full length CAT), but the big thing in this problem is you DO NOT account for order TWICE
the 4! is for the team
and the 2! is within every single combination for each pair

Yes, because in this question, the groups are not distinct. You have to split 8 people in 4 groups.
You can split them like this: (A, B), (C, D), (E, F), (G, H)
or like this: (G, H), (A, B), (C, D), (E, F)
they are the same split. They are not assigned to group1, group2, group3, group4.
Say the total number of ways we get = N

Now, if we change the question and say that we have 8 people and we need to divide them into 4 teams: Team 1, Team 2, Team 3 and Team 4

Then, one split is this: Team 1 = (A, B); Team 2 = (C, D), Team 3 = (E, F), Team 4 = (G, H)
and another split is: Team 1 = (G, H), Team 2 = (A, B); Team 3 = (C, D), Team 4 = (E, F)

These two cases were the same in our original question but if the teams/groups are distinct, the two cases are distinct. Now, the total number of ways = N*4!
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Re: combination (groups and that stuff...)  [#permalink]

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18 Nov 2013, 17:36
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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: In how many different ways can a group of 8 people be  [#permalink]

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17 Jul 2017, 03:54
manalq8 wrote:
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

Let's select the 4 groups of 2 individuals each

8C2 * 6C2 * 4C2 * 2C2

But wait... There is some problem...

Let 8 individuals are
A B C D E F G H

Case 1: selected 4 groups of 2 are respectively

A&B C&D E&F G&H

i.e. A&B us first group, C&D is second group etc

Case 2: selected 4 groups of 2 are respectively

C&D A&B G&H E&F respectively

So case 1 and 2 are the same except that order of the groups has been accounted for

The arrangements of 4 groups can be done in 4! Which needs to be excluded from calculation

Hence divide the result by 4!

So final result = 8C2 * 6C2 * 4C2 * 2C2 / 4! = 105

Hope this helps!!!
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Re: In how many different ways can a group of 8 people be  [#permalink]

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30 Aug 2017, 06:22
Bunuel wrote:
manalq8 wrote:
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

$$\frac{C^2_8*C^2_6*C^2_4*C^2_2}{4!}=105$$, we are dividing by 4! (factorial of the # of teams) as the order of the teams does not matter. If 8 people are - 1, 2, 3, 4, 5, 6, 7, 8, then (1,2)(3,4)(5,6)(7,8) would be the same 4 teams as (5,6)(7,8)(1,2)(3,4), as we don't have team #1, team #2...

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 left.

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

You can check similar problems:
http://gmatclub.com/forum/probability-8 ... %20equally
http://gmatclub.com/forum/probability-8 ... ide+groups
http://gmatclub.com/forum/combination-5 ... ml#p690842
http://gmatclub.com/forum/sub-committee ... ide+groups

There is also direct formula for this:

1. 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 is $$\frac{(mn)!}{(n!)^m*m!}$$.

2. 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 is $$\frac{(mn)!}{(n!)^m}$$

Hope it helps.

How did you use the formula? I did not get the answer using formula.
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In how many different ways can a group of 8 people be  [#permalink]

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30 Aug 2017, 09:45
gayatriv21 wrote:
Bunuel wrote:
manalq8 wrote:
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

$$\frac{C^2_8*C^2_6*C^2_4*C^2_2}{4!}=105$$, we are dividing by 4! (factorial of the # of teams) as the order of the teams does not matter. If 8 people are - 1, 2, 3, 4, 5, 6, 7, 8, then (1,2)(3,4)(5,6)(7,8) would be the same 4 teams as (5,6)(7,8)(1,2)(3,4), as we don't have team #1, team #2...

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 left.

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

You can check similar problems:
http://gmatclub.com/forum/probability-8 ... %20equally
http://gmatclub.com/forum/probability-8 ... ide+groups
http://gmatclub.com/forum/combination-5 ... ml#p690842
http://gmatclub.com/forum/sub-committee ... ide+groups

There is also direct formula for this:

1. 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 is $$\frac{(mn)!}{(n!)^m*m!}$$.

2. 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 is $$\frac{(mn)!}{(n!)^m}$$

Hope it helps.

How did you use the formula? I did not get the answer using formula.

We are dividing mn = 8 people into m = 4 teams of n = 2 people each.

Use the first formula, sine the order does not matter: $$\frac{(mn)!}{(n!)^m*m!}=\frac{8!}{(2!)^4*4!}=105$$.

P.S. You don;t really to memorize this formula for the GMAT.
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In how many different ways can a group of 8 people be divide  [#permalink]

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Updated on: 22 Aug 2019, 03:22
noboru wrote:
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

Number of ways can a group of 8 people be divided into 4 teams of 2 people each = $$^8C_2*^6C_2*^4C_2*^2C_2/4!$$ = 28*15*6*1 /4! = 105

IMO B
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Originally posted by Kinshook on 22 Aug 2019, 03:15.
Last edited by Kinshook on 22 Aug 2019, 03:22, edited 1 time in total.
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Re: In how many different ways can a group of 8 people be divide  [#permalink]

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22 Aug 2019, 03:21
noboru wrote:
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

Number of ways can a group of 8 people be divided into 4 teams of 2 people each = $$\frac{^8C_2*^6C_2*^4C_2*^2C_2}{4!} = \frac{28*15*6*1}{4!} = 105$$

IMO B
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Re: In how many different ways can a group of 8 people be divide   [#permalink] 22 Aug 2019, 03:21

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