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

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

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New post 22 Feb 2020, 05:18
Dear experts,
What if we had, FOR EXAMPLE, 3 teams of 1 individual in each and 1 team let's say consisting of 5 people. Will the answer be:
8C1*7C1*6C1*5C5/(4!/3!) ? Since we still have to unarrange , but have 3 repeating values.
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Re: In how many different ways can a group of 8 people be divide  [#permalink]

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New post 02 Mar 2020, 20:35
2
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


Responding to a pm:

Quote:
What if we had, FOR EXAMPLE, 3 teams of 1 individual in each and 1 team let's say consisting of 5 people. Will the answer be:
8C1*7C1*6C1*5C5/(4!/3!) ? Since we still have to unarrange , but have 3 repeating values. I know from your post that if all four teams had different number of people we would not unarrange. But if some repeat?


Yes, consider various scenarios:

8 people and 2 teams (one of 3 people and the other of 5 people)
You just pick 3 of the 8 people for the 3 people team. Rest everyone will be in the 5 people team. No un-arranging required.

8 people and 2 teams (one of 4 people and the other of 4 people)
You pick 4 people for the first team (say A, B, C, D). The other 4 (E, F, G, H) belong to the other team of 4 people. So you calculate this as 8C4. Now within this 8C4 will lie the case in which you picked (E, F, G, H) for the first team and ( A, B, C, D) will belong to the other team. But note that this case is exactly the same as before. 2 teams one (A, B, C, D) and other (E, F, G, H). So you need to un-arrange here.


Similarly, take your case - 8 people - 3 teams of 1 individual in each and 1 team let's say consisting of 5 people
We select 5 people out of 8 for the 5 people team in 8C5 ways. Rest of the 3 people play individually and we need to create no teams so nothing to be done. Since we are not doing any selection for a team, we are not inadvertently arranging and hence un-arranging is not required.

Instead say we have 8 people - and we make 3 teams, one with 4 people and two teams with 2 people each
We select 4 people out of 8 for the 4 people team in 8C4 ways.
Then we select 2 people for the first two people team in 4C2 ways and the leftover 2 people are for the second two people team. But again, as discussed above, there will same cases counted twice here so we will need to un-arrange by dividing by 2.
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In how many different ways can a group of 8 people be divide  [#permalink]

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New post 15 Mar 2020, 00:16
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.

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

Answer: B.

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.


Hi Bunuel VeritasKarishma IanStewart
we say we need to divide by 4! cus the order doesn't matter but when we use combinations arent we anyway implying that the order doesn't matter. I am not able to understand why we need to divide by 4! again when were using combinations and not permutations.

The second approach, I understand perfectly hence, I know I am missing something in the first one. Pls, help. This is really a confusing sub-topic for me under PnC
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In how many different ways can a group of 8 people be divide   [#permalink] 15 Mar 2020, 00:16

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