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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...

You can think about this in another way. 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.

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 \(\frac{(mn)!}{(n!)^m}\)

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

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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01 Mar 2013, 22:05

1

This post received KUDOS

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

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23 Jun 2014, 11:56

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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

Just think about this as you would any other combination problem.

You need to choose 4 groups of 2 from 8 people: That's \(\frac{8!}{2!2!2!2!}\), then divided by \(4!\), since any ordering of these 4 groups is the same.

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

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01 Mar 2013, 07:07

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, 21:09

I might be able to help you with this one.

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?

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

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03 Mar 2013, 06:31

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
_________________

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

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05 Mar 2013, 19:02

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

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
_________________

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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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

Answer Option B

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, 07:07

Hi all,

here you wants to divide 8 people in group of 4 teams in 2 members . So you can do these directly

8!/2!*2!*2!*2!*4!

so in these in the numerator denotes the total number of people and four times 2! denotes the four group and 4! denotes as the all the group has same number.

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

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30 Aug 2017, 07:22

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...

You can think about this in another way. 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.

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.

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...

You can think about this in another way. 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.

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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