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

Question

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

Bunuel · Accepted Answer

Official Solution:

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

A group of 8 people can be divided into 4 teams of 2 people each in  \frac{C^2_8*C^2_6*C^2_4*C^2_2}{4!}=\frac{28*15*6*1}{24}=105  different ways.
 
Above,  C^2_8  is the number of ways to choose 2 people for the first pair,  C^2_6  is the number of ways to choose 2 people for the second pair,  C^2_4  is the number of ways to choose 2 people for the third pair, and  C^2_2  is the number of ways to choose 2 people for the fourth pair. 
 
We divide by 4! (the factorial of the number of pairs) because the order of the pairs does not matter. For example, if the 8 people are labeled 1, 2, 3, 4, 5, 6, 7, and 8, the pair-selection (1,2); (3,4); (5,6); (7,8) would be considered the same as (7,8); (1,2); (3,4); (5,6), as there is no specific order assigned to the pairs. And since  C^2_8*C^2_6*C^2_4*C^2_2  will give all 4! arrangements of (1,2); (3,4); (5,6); (7,8) pair-selection, we need to divide this and each other possible pair-selection by 4! to eliminate the essentially identical pair-selections.
 
An alternative way to think about this is as follows:
   
• For the first person, there are 7 possible choices for a teammate;
 
• For the next unpaired person, there are 5 possible choices for a teammate (as two people are already chosen);
 
• For the third unpaired person, there are 3 possible choices for a teammate (as four people are already chosen);
 
• For the fourth unpaired person, there is only one choice left for a teammate.
  
Multiplying these choices together gives us 7*5*3*1 = 105 different ways to divide the group of 8 people into 4 teams of 2 people each.

Answer: B

For similar questions check  DISTRIBUTING ITEMS/PEOPLE/NUMBERS... (QUESTION COLLECTION)

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.

Snowbirds · Answer

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

IanStewart · Answer

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.

dandarth1 · Answer

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)?

Bunuel · Answer

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.

manimgoindown · Answer

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

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

KarishmaB · Answer

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.

manimgoindown · Answer

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

KarishmaB · Answer

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!

GMATinsight · Answer

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

GMATinsight · Answer

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

GMATinsight · Answer

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!!!

Kinshook · Answer

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

KarishmaB · Answer

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.

Vordhosbn · Answer

- (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.

GMAT2point0 · Answer

Hey! I had a follow up question for the new cases you shared.

I understood case 1 (that when making one team of 5 ppl and 3 teams of 1 person each, we don't need to divide 8C5 by anything as we haven't created any ordering double calculations here)

However, in case 2 (when creating 3 teams, one of 4 ppl and 2 of 2 ppl each), why will we divide only by 2!. I thought we will need to divide by 3! as 8C4 * 4C2 * 2C2 would introduce multiple occurrences of the same set of teams. Is there something I can't imagine here?

Please let me know. Thanks!

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

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