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

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Intern
Joined: 20 Oct 2019
Posts: 2
Re: In how many different ways can a group of 8 people be divide  [#permalink]

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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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02 Mar 2020, 20:35
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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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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...

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.

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
In how many different ways can a group of 8 people be divide   [#permalink] 15 Mar 2020, 00:16

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