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# M06-11

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Math Expert
Joined: 02 Sep 2009
Posts: 44600

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16 Sep 2014, 00:27
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Difficulty:

95% (hard)

Question Stats:

31% (00:58) correct 69% (01:30) wrong based on 137 sessions

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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
[Reveal] Spoiler: OA

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Math Expert
Joined: 02 Sep 2009
Posts: 44600

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16 Sep 2014, 00:27
1
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Expert's post
12
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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 $$\frac{C_8^2*C_6^2*C_4^2*C_2^2}{4!}=105$$.

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Joined: 11 Sep 2013
Posts: 151
Concentration: Finance, Finance

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08 Oct 2014, 01:26
If the question were"order doesn't matter", would the answer be 2C8/4!?

Please explain. What is the issue of order here?
Math Expert
Joined: 02 Sep 2009
Posts: 44600

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08 Oct 2014, 05:03
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Raihanuddin wrote:
If the question were"order doesn't matter", would the answer be 2C8/4!?

Please explain. What is the issue of order here?

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...

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

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Hope this helps.
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08 Oct 2014, 11:01
Thank you
Intern
Joined: 19 Nov 2013
Posts: 25
Location: India
Concentration: Strategy, Technology
WE: Information Technology (Computer Software)

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22 Oct 2014, 21:27
I am getting E

8!/(8/4!)^4=2520
Intern
Joined: 17 Mar 2014
Posts: 28
Location: United States
Concentration: Strategy, General Management
GMAT 1: 710 Q49 V36
WE: Engineering (Energy and Utilities)

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06 Nov 2014, 07:00
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
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Joined: 28 Oct 2014
Posts: 11
GMAT 1: 710 Q47 V40

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27 Dec 2014, 20:19
1
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Bunuel wrote:
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 $$\frac{C_8^2*C_6^2*C_4^2*C_2^2}{4!}=105$$.

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.

Current Student
Joined: 18 Sep 2015
Posts: 95
GMAT 1: 610 Q43 V31
GMAT 2: 610 Q47 V27
GMAT 3: 650 Q48 V31
GMAT 4: 700 Q49 V35
WE: Project Management (Health Care)

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14 Apr 2016, 12:25
2
KUDOS
- 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.
Intern
Joined: 12 May 2017
Posts: 8

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28 Dec 2017, 01:13
Bunuel wrote:
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 $$\frac{C_8^2*C_6^2*C_4^2*C_2^2}{4!}=105$$.

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! ?
Math Expert
Joined: 02 Sep 2009
Posts: 44600

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28 Dec 2017, 01:20
Deepshikha1907 wrote:
Bunuel wrote:
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 $$\frac{C_8^2*C_6^2*C_4^2*C_2^2}{4!}=105$$.

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
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Math Expert
Joined: 02 Sep 2009
Posts: 44600

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28 Dec 2017, 01:23
Deepshikha1907 wrote:
Bunuel wrote:
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 $$\frac{C_8^2*C_6^2*C_4^2*C_2^2}{4!}=105$$.

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! ?

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Re: M06-11   [#permalink] 28 Dec 2017, 01:23
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# M06-11

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