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

Another way to think about this:

How many ways can you arrange the following:
T1 T1 T2 T2 T3 T3 T4 T4

That would be: 8!/(2!*2!*2!*2!)

Then also recall that we don't care about the differences between the teams, therefore

8!/(2!*2!*2!*2!*4!) = 105

I understand how to get the answer, but I'm wondering why we assume that the order of the teams doesn't matter. The question just asks how many ways you can group them. When should we assume something matters or doesn't matter when the question doesn't specify? And also, just to clarify, when we do 8!/2!2!2!2!4!, we are also not worrying about the order within each team either right i.e. (a,b)=(b,a)?

The teams are not numbered/labeled (we don't have team #1, #2, ...), the teams are not assigned to something (for example to tournaments), ... So, the order of the teams doesn't matter.

Please check the links in my previous post for similar problems.

Hope it helps.

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

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!

Ways of choosing 2 out od a group of n people = nC2

i.e. Ways of choosing first team of 2 out of 8 = 8C2
i.e. Ways of choosing Second team of 2 out of remaining 6 = 6C2
i.e. Ways of choosing Third team of 2 out of remaining 4 = 4C2
Remaining 2 will form Forth team

Total Ways of Choosing Teams = 8C2 * 6C2 * 4C2 * 1

BUT Since the first team may come on second places and second may come on third etc. i.e.e arrangement among teams are included here which we NEED to exclude

i.e. Total Ways of Choosing Teams = 8C2 * 6C2 * 4C2 * 1 / 4! = 28*15*6/24 = 105

Answer: option B

The FIRST pair can be chosen in 8C2 ways

The SECOND pair can be chosen in 6C2 ways as 6 members are left after choosing 1st pair

The THIRD pair can be chosen in 4C2 ways as 4 members are left after choosing 1st two pair and so on...

Therefore total ways of splitting 8 members in 6 pairs = (8C2 * 6C2 * 4C2 * 2C2) / 4!


The entire expression is divided by 4! because the arrangements of pair have been accounted for, which need to be excluded. Arrangement for example Same pair which came at first place can be chosen at third place or forth places also while we are choosing different pairs at different places

Required answer = (28 * 15 * 6 * 1)/24 = 105

Answer: option C

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!!!

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

Responding to a pm:



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.

- (1) Captain from the first team has 7 options (excluding the captain).
- (2) Two people are gone from (1), and we need to exclude the captain from the second team, giving us 6-2-1 = 5. Thus, 5 options left for the second captain.
- (3) Four people are gone, Excluding the captain from the third team (8-4-1) we have 3 options left for the last captain.
- (4) 7 * 5 * 3 =105.

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