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

\(\frac{C^2_8*C^2_6*C^2_4*C^2_2}{4!}=105\), 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.

Similar questions to practice:
nine-dogs-are-split-into-3-groups-to-pull-one-of-three-88685.html
in-how-many-different-ways-can-a-group-of-9-people-be-85993.html
a-group-of-8-friends-want-to-play-doubles-tennis-how-many-55369.html
in-how-many-different-ways-can-a-group-of-8-people-be-divide-85707.html
in-how-many-different-ways-can-a-group-of-8-people-be-99053.html
in-how-many-different-ways-can-a-group-of-9-people-be-divide-101722.html

Hope this helps.
_________________

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

The solution to this problem is the number of combinations. First we get one team out of 8 . The number of ways to do this would be \(C_8^2\). The next combination is 2 out of 6 or \(C_6^2\), and so on. Having all four combinations multiplied, we need to divide the total number by the number of ways the teams can be chosen \(4!\), since we are not interested if the team with two certain people is chosen first, second or third. Therefore, the answer is found by the following formula \(\frac{C_8^2*C_6^2*C_4^2*C_2^2}{4!}=105\).

Answer: B
_________________

If the question were"order doesn't matter", would the answer be 2C8/4!?

Please explain. What is the issue of order here?

Thank you

I am getting E

8!/(8/4!)^4=2520

okay i didn't really get it and i don't expect anyone to explain . but my Gmat test is on sunday and i would like someone to give me a rule here if possible.
this division by 4! reminded me of something we learned at school when studying probability and i remember there was a rule about how to identify questions where you calculate a numerator using a combination and then divide by a factorial , there is a certain type of questions that requires this "unique approach" , does anyone know how to identify this type of questions?
i don't want a general explanation about combinations because i know them , just how to identify a question where in the answer you need to divide by a factorial .

thanks

Hi,

I'm having a difficult time conceptualizing the explanation to this problem. I'm not good with formal notation for combination problems so I use the simple "anagram" method, which I learned from MGMAT. If anyone is familiar with that approach, could you please post an explanation to the problem using that language?

In approaching this problem, my thought process was something similar to what follows below.

8 people into 4 teams of 2... Okay 8 choose 2, 4 different ways. So I did an anagram with 1, 2, 3, ...8 and A, A, B, B, ... H, H, something I represented mathematically with 8! / 2!*2!*2!*2!. Then I thought about how many ways I could make 4 teams out of that and divided the result, which was 2520, by 4! and got the answer, 105.

I feel like my approach was very lucky in this situation and won't work well in the future.

Thanks for your time.

Another way to think about this is remove the:
- meaning for order within the groups
- meaning for order among the groups
solution:
1. we arrange all the 8 people in a line, this gives us: 8!
2. since the order among the gourps does not matter, we will have to devide 8! by 4!
3. since the order within the groups allso does not matter, for each group, we will have to devide by 2!

hence we got: 8!/(4!2!2!2!2!) = 105

comment: by saying "the order does not matter" i mean that, for example if we take the order within the gourp, it does matter if i picked person X to be the first in the group and later person Y to be in the group.

Bunuel, can you please explain why the total no. of ways is divided by 4! when we are not interested if the team with two certain people is chosen first, second or third? I mean how have you determined that it should be divided by 4! ?

Check here: https://gmatclub.com/forum/m06-183708.html#p1425150
_________________

DISTRIBUTING ITEMS/PEOPLE/NUMBERS... (QUESTION COLLECTION):

https://gmatclub.com/forum/in-how-many- ... 87128.html
https://gmatclub.com/forum/in-how-many- ... 25669.html
https://gmatclub.com/forum/larry-michae ... 08739.html
https://gmatclub.com/forum/in-how-many- ... 26991.html
https://gmatclub.com/forum/in-how-many- ... 98697.html
https://gmatclub.com/forum/how-many-pos ... 85291.html
https://gmatclub.com/forum/how-many-way ... 37198.html
https://gmatclub.com/forum/in-how-many- ... 41072.html
https://gmatclub.com/forum/mrs-smith-ha ... 98225.html
https://gmatclub.com/forum/in-how-many- ... 70689.html
https://gmatclub.com/forum/in-how-many- ... 05384.html
https://gmatclub.com/forum/four-boys-pi ... 98701.html
https://gmatclub.com/forum/how-many-way ... 61598.html
https://gmatclub.com/forum/in-how-many- ... 73423.html
http://gmatclub.com/forum/in-how-many-w ... 41070.html
https://gmatclub.com/forum/in-how-many- ... 64389.html
https://gmatclub.com/forum/in-how-many- ... 33322.html
https://gmatclub.com/forum/in-how-many- ... 31187.html
https://gmatclub.com/forum/in-how-many- ... 06279.html
https://gmatclub.com/forum/in-how-many- ... 26348.html
https://gmatclub.com/forum/in-how-many- ... 81816.html
https://gmatclub.com/forum/in-how-many- ... 34223.html
https://gmatclub.com/forum/in-how-many- ... 40328.html
https://gmatclub.com/forum/in-how-many- ... 40329.html
https://gmatclub.com/forum/mel-and-nora ... 34163.html

HARD (FOR PRACTICE): https://gmatclub.com/forum/5-rings-on-4 ... 86111.html
_________________

To combine the two approaches. I guess by solving to get 2520, we are doing stage 1 and 3 of your approach. Dividing by 4! is the stage 2...

Hi Bunuel,

Why can't we solve this question like this:

The first person can be place among four different teams,
The second person also among four different teams,
The third, among 3,
The fourth among 3,
The fifth between 2,
The sixth between 2,
The seventh and eighth have no option.
Therefore the total number of ways = 4x4x3x3x2x2 =576

Thanks for clarifying
P

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

The solution to this problem is the number of combinations. First we get one team out of 8 . The number of ways to do this would be C28C82. The next combination is 2 out of 6 or C26C62, and so on. Having all four combinations multiplied, we need to divide the total number by the number of ways the teams can be chosen 4!4!, since we are not interested if the team with two certain people is chosen first, second or third. Therefore, the answer is found by the following formula C28∗C26∗C24∗C224!=105C82∗C62∗C42∗C224!=105.


@Experts --
Division by 4! , CAN YOU RELATE THIS TO PERMUTATION AND COMBINATION. Not able to understand that while using Permutation we usually take care of Order / Arrangement but here while applying Combination we are dividing by 4! to undo arrangement. Confused !!
_________________

Hello guys I was wondering if someone could point out the flaw in my approach

(8*7/2!) * (6*5/2!) * (4*3/2!) * (2*1/2!)=2520

there has to be 1 more step that I'm missing or the whole approach is wrong.

We need to make 4 teams of 2 people of out 8 people. Here, we will need to select 2 out of 8 and then 2 out of remaining 6, then 2 out of remaining 4 and then 2 out of remaining 2.

Notice that the order in which the teams are selected does not matter.

So, this is a permutation of combinations. Since there are 4 teams and the order in which they are selected does not matter, we divide by 4! to remove the effect of arrangement.

Permutation/arrangement = Combination

Posted from my mobile device

Here you need to divide the answer by 4! that is 24, as order among the teams doesn't matter. So you will get the final answer that is 105

I had trouble conceptualizing this problem at first, and after re-visiting Combinatorics theory and Bunuel's response I realized the best way I understood it was as having 2 sets of combination selections: 1) Selecting players per each team of 2, 2. Selecting 4 teams. Hence that is why the Combination formula needs to be applied twice.

Illustrating this:

1) Selecting players per each team of 2:

Out of 8 players, selecting 1st team:
\(\frac{8!}{(8-2!)2!}\) = 28 or else \(\frac{8*7}{2!}\) = 28

Since we have selected 2 out of 8 players, that leaves 6 possible players for 2nd team:
\(\frac{6*5}{2!}\) = 15

3rd team:
\(\frac{4*3}{2!}\) = 6

4th team:
\(\frac{2*1}{2!}\) = 1

2) Selecting 4 teams (Combination instead of Permutation since order does not matter):

\(\frac{(1st team * 2nd team* 3rd team * 4th team)}{4!}\)

= \(\frac{(28*15*6*1)}{4!}\)

= 105

Hi Bunuel,

Could you please explain why we are dividing by 4! when we are using the combinations formula, isn't the combinations formula used only for selection?
There shouldn't be any arrangement involved if it's using that. Please tell me where i'm faltering here.

Thanks