Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 11 Feb 2016, 14:05

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# m06Q11

 Question banks Downloads My Bookmarks Reviews Important topics
Author Message
Senior Manager
Joined: 12 Mar 2009
Posts: 312
Followers: 1

Kudos [?]: 183 [0], given: 1

m06Q11 [#permalink]  04 Jul 2009, 06:37
6
This post was
BOOKMARKED
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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
B

Source: GMAT Club Tests - hardest GMAT questions

pl explain the ans particularly usage of 4!.
 Kaplan GMAT Prep Discount Codes Knewton GMAT Discount Codes GMAT Pill GMAT Discount Codes
Manager
Joined: 28 Jan 2004
Posts: 203
Location: India
Followers: 2

Kudos [?]: 22 [1] , given: 4

Re: m06Q11 [#permalink]  04 Jul 2009, 14:10
1
KUDOS
May be someone with a good understanding of permutations and combinations can explain this discrepency.
Personally I never doubt the explanations given by AkamaiBrah but i dont know about this particular question.
May be mods can help.
Manager
Joined: 11 Apr 2009
Posts: 211
Followers: 3

Kudos [?]: 105 [5] , given: 4

Re: m06Q11 [#permalink]  05 Jul 2009, 01:54
5
KUDOS
vaivish1723 wrote:
pl explain the ans particularly usage of 4!.

By multiplying all the combinations of different groups you are getting a figure for which order matters

For example if there are 8 people: john,mary,jane,rima,sandeep,huhu,bruce and ram

the groups (john,mary)(jane,rima)(sandeep,huhu)(bruce,ram) is the same as (bruce,ram)(john,mary)(sandeep,huhu)(jane,rima)

as mentioned earlier by multiplying the number of various combinations possible you get all the different ways the groups could be organised without differentiating that order does not patter. The way to discount this is to divide by the factorial of the number of items = 4!

I hope that makes sense

((28)(15)(6)(1))/(4)(3)(2) = 105.
_________________

-talent is the desire to practice-

Intern
Joined: 26 Jun 2010
Posts: 3
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: m06Q11 [#permalink]  25 Oct 2010, 04:26
Could you please explain how you came up with each number: ((28)(15)(6)(1))/(4)(3)(2) = 105.

I am confused. Would be really helpful.
Intern
Joined: 25 Aug 2010
Posts: 46
Followers: 1

Kudos [?]: 12 [2] , given: 3

Re: m06Q11 [#permalink]  25 Oct 2010, 05:02
2
KUDOS
ileannaconst wrote:
Could you please explain how you came up with each number: ((28)(15)(6)(1))/(4)(3)(2) = 105.

I am confused. Would be really helpful.

get 2 people from 8 = $$\frac{8!}{6!*2!}$$=28
get 2 people from 8-2 = $$\frac{6!}{4!*2!}$$=15
get 2 people from 8-4 = $$\frac{4!}{2!*2!}$$=6
get 2 people from 8-6 = $$\frac{2!}{0!*2!}$$=1
Senior Manager
Joined: 25 Jul 2010
Posts: 300
Location: India
Concentration: Strategy, Technology
GMAT 1: 770 Q51 V46
GPA: 3.2
WE: Engineering (Computer Software)
Followers: 19

Kudos [?]: 31 [4] , given: 20

Re: m06Q11 [#permalink]  25 Oct 2010, 05:11
4
KUDOS
Hi,
I didn't know that we had to divide by 4!, I just knew that we had to divide by some number to make up for the repetitions. So I didn't know what to do with the 840 I kept on getting.
So I left PnC and just tried another logic:

Here it is..

I took up a smaller group of 4 persons, whom I have to divide into 2 teams - A B C D

So the 1st team could have any one of AB, AC or AD - 3 and the other team in each of the cases has only one option - the other 2 persons.
-- So it becomes 3 x 1
When I extended it to 6 persons into 2 teams, I got - 5 x 3 x 1

So I extended till 8 and got it 7 x 5 x 3 x 1 = 105
Manager
Status: I rest, I rust.
Joined: 04 Oct 2010
Posts: 122
Schools: ISB - Co 2013
WE 1: IT Professional since 2006
Followers: 17

Kudos [?]: 108 [25] , given: 9

Re: m06Q11 [#permalink]  25 Oct 2010, 09:37
25
KUDOS
Here is another conceptual way:

Step1: Lets take one person from 8. Now in how many ways can he be teamed up with another man? 7!!! Team-1 is formed.
Step2: Lets take another person from remaining 6. In how many ways can he be teamed up with another man? 5!!! Team-2 is formed.
Step3: Lets take another person from remaining 4. In how many ways can he be teamed up with another man? 3!!! Team-3 is formed.
Step4: Last two can be teamed up in only 1 way.

So total ways = 7*5*3*1 = 105
_________________

Respect,
Vaibhav

PS: Correct me if I am wrong.

Manager
Affiliations: The Earth organization, India
Joined: 25 Dec 2010
Posts: 193
WE 1: SAP consultant-IT 2 years
WE 2: Entrepreneur-family business 2 years
Followers: 5

Kudos [?]: 11 [0], given: 12

Re: m06Q11 [#permalink]  15 Jun 2011, 07:46
if we dont go by the brilliant method suggested by srijaidev and instead go with conventional combinatorics as suggested by nightwing:

WHAT SHOULD BE THE WORDING OF THE QUESTION SO THAT WE DO NOT NEED TO DIVIDE THE TOTAL NUMBER OF COMBINATIONS BY 4! ?

I am posting this query so that I can understand the difference in the 2 question types if faced by either on G-day.
_________________

Cheers !!

Quant 47-Striving for 50
Verbal 34-Striving for 40

Manager
Status: livin on a prayer!!
Joined: 12 May 2011
Posts: 124
Location: Australia
Followers: 0

Kudos [?]: 21 [2] , given: 1

Re: m06Q11 [#permalink]  04 Nov 2011, 00:39
2
KUDOS
In!/Out!

8!/(4!2!2!2!2!)
= 7 * 5 * 3
= 105
_________________

Aim for the sky! (800 in this case)
If you like my post, please give me Kudos

Senior Manager
Status: mba here i come!
Joined: 07 Aug 2011
Posts: 270
Location: Pakistan
Concentration: Strategy, Marketing
GMAT 1: 680 Q46 V37
GMAT 2: Q V
Followers: 38

Kudos [?]: 894 [1] , given: 48

Re: m06Q11 [#permalink]  26 Mar 2012, 05:58
1
KUDOS
$$\frac{1}{4!}*[\frac{8*7}{2!}*\frac{6*5}{2!}*\frac{4*3}{2!}*\frac{2*1}{2!}] = 105$$

$$\frac{1}{4!}$$ because order of groups does not matter.
_________________

press +1 Kudos to appreciate posts

Manager
Status: I will not stop until i realise my goal which is my dream too
Joined: 25 Feb 2010
Posts: 235
Schools: Johnson '15
Followers: 2

Kudos [?]: 42 [1] , given: 16

Re: m06Q11 [#permalink]  26 Mar 2012, 06:20
1
KUDOS
vaibhavtripathi wrote:
Here is another conceptual way:

Step1: Lets take one person from 8. Now in how many ways can he be teamed up with another man? 7!!! Team-1 is formed.
Step2: Lets take another person from remaining 6. In how many ways can he be teamed up with another man? 5!!! Team-2 is formed.
Step3: Lets take another person from remaining 4. In how many ways can he be teamed up with another man? 3!!! Team-3 is formed.
Step4: Last two can be teamed up in only 1 way.

So total ways = 7*5*3*1 = 105

For poor chaps like me who find the permutation, combination, probablity as a demon, your answers were simple to understand buddy....

+1 to u
_________________

Regards,
Harsha

Note: Give me kudos if my approach is right , else help me understand where i am missing.. I want to bell the GMAT Cat

Satyameva Jayate - Truth alone triumphs

Math Expert
Joined: 02 Sep 2009
Posts: 31297
Followers: 5358

Kudos [?]: 62425 [7] , given: 9455

Re: m06Q11 [#permalink]  26 Mar 2012, 06:46
7
KUDOS
Expert's post
vaivish1723 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

[Reveal] Spoiler: OA
B

Source: GMAT Club Tests - hardest GMAT questions

pl explain the ans particularly usage of 4!.

$$\frac{C^2_8*C^2_6*C^2_4*C^2_2}{4!}=105$$, 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:
probability-88685.html?hilit=different%20items%20divided%20equally
probability-85993.html?highlight=divide+groups
combination-55369.html#p690842
sub-committee-86346.html?highlight=divide+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.
_________________
Intern
Joined: 08 May 2012
Posts: 2
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: m06Q11 [#permalink]  07 Jun 2012, 13:59
Hi,

I still have some trouble to figure out when to account for repetitions . I thought that by calculating the number of possibilities for the first, second, third and fourth team, I create 4 different and independent "pools" of choices.
To me, the problem seemed similar to the following:

9 basketball players are trying out to be on a newly formed basketball team. Of these players, 5 will be chosen for the team. If 6 of the players are guards and 3 of the players are forwards, how many different teams of 3 guards and 2 forwards can be chosen?
(OA is 60 - first step: calculate possibilities for the two different pools - second step: multiply by each other WITHOUT accounting for repetitions.

It would be great if anyone could help me on this, it drives me crazy

Thank you so much

Elzi
Manager
Joined: 01 May 2012
Posts: 57
GPA: 1
Followers: 2

Kudos [?]: 7 [0], given: 0

Re: m06Q11 [#permalink]  07 Jun 2012, 21:59
What's the probability that combinatorics appear on the exam?
Math Expert
Joined: 02 Sep 2009
Posts: 31297
Followers: 5358

Kudos [?]: 62425 [1] , given: 9455

Re: m06Q11 [#permalink]  08 Jun 2012, 02:47
1
KUDOS
Expert's post
Elzi wrote:
Hi,

I still have some trouble to figure out when to account for repetitions . I thought that by calculating the number of possibilities for the first, second, third and fourth team, I create 4 different and independent "pools" of choices.
To me, the problem seemed similar to the following:

9 basketball players are trying out to be on a newly formed basketball team. Of these players, 5 will be chosen for the team. If 6 of the players are guards and 3 of the players are forwards, how many different teams of 3 guards and 2 forwards can be chosen?
(OA is 60 - first step: calculate possibilities for the two different pools - second step: multiply by each other WITHOUT accounting for repetitions.

It would be great if anyone could help me on this, it drives me crazy

Thank you so much

Elzi

This concept is explained here:
probability-88685.html?hilit=different%20items%20divided%20equally
probability-85993.html?highlight=divide+groups
combination-55369.html#p690842
sub-committee-86346.html?highlight=divide+groups
_________________
Intern
Joined: 08 May 2012
Posts: 2
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: m06Q11 [#permalink]  08 Jun 2012, 14:42
Thanks a lot Bunuel!
Intern
Joined: 09 Jun 2012
Posts: 11
Location: United Kingdom
Schools: LBS MIF '14
GMAT 1: Q V0
WE: Business Development (Investment Banking)
Followers: 0

Kudos [?]: 5 [0], given: 3

Re: m06Q11 [#permalink]  29 Oct 2012, 10:01
Another option is to use the formula (found it in one of the threads):

(m*n)!/[(n!)^m]*m!
m - number of people
n - number of groups

8!/[(2)^4]*4! = 105
Math Expert
Joined: 02 Sep 2009
Posts: 31297
Followers: 5358

Kudos [?]: 62425 [0], given: 9455

Re: m06Q11 [#permalink]  29 Oct 2012, 10:06
Expert's post
EV wrote:
Another option is to use the formula (found it in one of the threads):

(m*n)!/[(n!)^m]*m!
m - number of people
n - number of groups

8!/[(2)^4]*4! = 105

That's correct. Check here for the formula: m06q11-80504.html#p1065072
_________________
Intern
Joined: 30 Aug 2012
Posts: 9
Concentration: General Management, Finance
Followers: 0

Kudos [?]: 7 [0], given: 3

Re: m06Q11 [#permalink]  03 Nov 2012, 00:44
Simplify the question in this way, you need to select a team of two from 8 = 8c2 = 28
now 6 are left and = 6c2 = 15 4c2 = 6 and finally only two people left =1

But you can select these in 4! ways.
so final selection will be (8c2 * 6c2 * 4c2 * 2c2) / (4 !) = 105
Current Student
Joined: 12 Dec 2012
Posts: 33
Concentration: Leadership, Social Entrepreneurship
GMAT 1: Q V
GMAT 2: 660 Q48 V33
GMAT 3: 740 Q49 V41
GPA: 3.74
Followers: 9

Kudos [?]: 89 [0], given: 19

Re: m06Q11 [#permalink]  25 Oct 2013, 23:55
Elzi wrote:
Hi,

I still have some trouble to figure out when to account for repetitions . I thought that by calculating the number of possibilities for the first, second, third and fourth team, I create 4 different and independent "pools" of choices.
To me, the problem seemed similar to the following:

9 basketball players are trying out to be on a newly formed basketball team. Of these players, 5 will be chosen for the team. If 6 of the players are guards and 3 of the players are forwards, how many different teams of 3 guards and 2 forwards can be chosen?
(OA is 60 - first step: calculate possibilities for the two different pools - second step: multiply by each other WITHOUT accounting for repetitions.

It would be great if anyone could help me on this, it drives me crazy

Thank you so much

Elzi

I think the trick is not to get too technical, but to rely on logic. Here's how I did that (and feel free to correct my thinking)

-I have 8 people and I need to figure out how many DIFFERENT ways I can split these guys into 4 teams of 2. I'm making a chain of 4 teams. A chain is different only when there's a new team in there, not a different arrangement of the same 4 teams!
-Let me call these teams T1, T2, T3, T4
-Clearly what the question is saying is that I don't care if I pick them the following ways, it's all the same to the author
T2 T1 T3 T4
T2 T3 T1 T4
and 22 other varieties (i.e. a total of 4! or 24 ways I can arrange these 4 SPECIFIC teams) - ok, noted
-Let's see how I can choose T1, T2, T3, T4
T1 - I can choose 8C2 ways
T2 - I can choose 6C2 ways
T3 - I can choose 4C2 ways
T4 - I can choose 2C2 or 1 way (since well there are only 2 guys left)

The key is NOT to get confused on the usage of combinations and assume that since we're using combinations the order already doesnt matter. Sure in the last team say only Jack and Jill were left - we still only picked one team - i.e. we didn't distinguish between (Jack-Jill) or (Jill-Jack). This doesnt mean that we are done. We need to focus on what the question is asking us and we need to give them DISTINCT ways of PICKING 4 TEAMS of 2 (i.e. this is one further layer)

8C2 * 6C2 * 4C2 * 2C2 = total number of ways I can arrange Teams 1, 2, 3 and 4. But I don't care about how these teams are arranged. Therefore, I divide by 4!

It helps to try and understand what the author is asking rather than get stuck on what formula to use.

I'm also trying to solidify my understanding of when to adjust for arrangements etc. so this is work-in-progress!

Hope this helps.
Re: m06Q11   [#permalink] 25 Oct 2013, 23:55

Go to page    1   2    Next  [ 21 posts ]

Display posts from previous: Sort by

# m06Q11

 Question banks Downloads My Bookmarks Reviews Important topics

Moderators: Bunuel, WoundedTiger

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.