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

 It is currently 22 Jul 2014, 07:35

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
Your Progress

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

How many ways are there to split a group of 6 boys into two

 Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:
Current Student
Joined: 11 May 2008
Posts: 560
Followers: 7

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

How many ways are there to split a group of 6 boys into two [#permalink]  26 Aug 2008, 09:20
1
This post was
BOOKMARKED
How many ways are there to split a group of 6 boys into two groups of 3 boys each? (The order of the groups does not matter)

* 8
* 10
* 16
* 20
* 24

why is the ans not 6c3?

suppose the quesiton was 3 groups of 2 boys each, what would be the ans?
Senior Manager
Joined: 06 Apr 2008
Posts: 452
Followers: 1

Kudos [?]: 40 [0], given: 1

Re: gmat club [#permalink]  26 Aug 2008, 09:31
arjtryarjtry wrote:
How many ways are there to split a group of 6 boys into two groups of 3 boys each? (The order of the groups does not matter)

* 8
* 10
* 16
* 20
* 24

why is the ans not 6c3?

suppose the quesiton was 3 groups of 2 boys each, what would be the ans?

Answer is 6C3 = 20

For 3 groups of 2 boys it will be 6C2*4C2*2C2 = 15*6 = 90
Manager
Joined: 15 Jul 2008
Posts: 210
Followers: 3

Kudos [?]: 21 [0], given: 0

Re: gmat club [#permalink]  26 Aug 2008, 09:32
arjtryarjtry wrote:
How many ways are there to split a group of 6 boys into two groups of 3 boys each? (The order of the groups does not matter)

* 8
* 10
* 16
* 20
* 24

why is the ans not 6c3?

suppose the quesiton was 3 groups of 2 boys each, what would be the ans?

The answer is 6C3/2! = 10

6C3 is right if you are asked to form one group of 3 people.
The actual expression is 6C3*3C3, where 3C3 =1.
But here you are bringing in order - group formed 1st , group formed 2nd. Since order does not matter, divide by the number of wasy the group can be arranged.

If you had to form 3 groups of two ppl each, then

first group can be formed in 6C2 ways. The second group can be formed in 4C2 ways and the third one can be formed in 2C2 ways.
so you have 6C2*4C2*2C2. But this orders the the three groups into 3! ways.
Your answer would be 6C2*4C2*2C2/3! = 15

The "C"s in the 6C2*4C2*2C2 eliminate the ordering of players only. Not the ordering of teams. You need to account for the ordering of teams also to find the true answer.

Hope that helps..

Last edited by bhushangiri on 26 Aug 2008, 09:41, edited 1 time in total.
Senior Manager
Joined: 06 Apr 2008
Posts: 452
Followers: 1

Kudos [?]: 40 [0], given: 1

Re: gmat club [#permalink]  26 Aug 2008, 09:40
bhushangiri wrote:
arjtryarjtry wrote:
How many ways are there to split a group of 6 boys into two groups of 3 boys each? (The order of the groups does not matter)

* 8
* 10
* 16
* 20
* 24

why is the ans not 6c3?

suppose the quesiton was 3 groups of 2 boys each, what would be the ans?

The answer is 6C3/2! = 10

6C3 is right if you are asked to form one group of 3 people.
The is because the actual expression is 6C3*3C3.
But here you are bringing in order - group formed 1st , group formed 2nd. Since order does not matter, divide by the number of wasy the group can be arranged.

If you had to form 3 groups of two ppl each, then

first group can be formed in 6C2 ways. The second group can be formed in 4C2 ways and the third one can be formed in 2C2 ways.
so you have 6C2*4C2*2C2. But this orders the the three groups into 3! ways.
Your answer would be 6C2*4C2*2C2/3! = 15

The "C"s in the 6C2*4C2*2C2 choose only the players. But not the teams. You need to account for the ordering of teams.

Hope that helps..

Right ... keep slipping out of my mind. Last time I calculated rightly , this time wrong
Current Student
Joined: 11 May 2008
Posts: 560
Followers: 7

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

Re: gmat club [#permalink]  26 Aug 2008, 09:52
thanks bhushan.. it helped me.
so we have to divide b the no. of grps thus formed, and multiplication should be done for the same no. of grps is it ?
eg
three grps -> 6c2*4c2*2c2/3!
two grps-> 6c3*3c3/3!

suppose it was 1 group of 4 people
is it just 6c4?
GMAT Instructor
Joined: 04 Jul 2006
Posts: 1270
Location: Madrid
Followers: 23

Kudos [?]: 112 [0], given: 0

Re: gmat club [#permalink]  26 Aug 2008, 09:53
Also, look from the tallest boy's perspective: 2 of the 5 other boys will be on his team: 5C2 = 10
SVP
Joined: 30 Apr 2008
Posts: 1893
Location: Oklahoma City
Schools: Hard Knocks
Followers: 28

Kudos [?]: 423 [0], given: 32

Re: gmat club [#permalink]  26 Aug 2008, 10:19
arjtryarjtry wrote:
thanks bhushan.. it helped me.
so we have to divide b the no. of grps thus formed, and multiplication should be done for the same no. of grps is it ?
eg
three grps -> 6c2*4c2*2c2/3!
two grps-> 6c3*3c3/3!

suppose it was 1 group of 4 people
is it just 6c4?

Yes, if it is C_6^4 for selecting 1 group of 4 people from 6. The thing that I forgot with regard to this question was the significance of "order does not matter." I was thinking that C_6^3 was correct, but didn't remember to divide by the permutations of the order of the group. For instance.

3 groups of 3 from 9 people

\frac{C_3^9 * C_3^6 * C_3^3}{3!}

The first group formed can have any 3 of the 9. next group will be any 3 of the 6 remaining, and the last group is formed by selection the 3 from the 6, because you have 3 remaining, which is why it will always be 1. But then you have 3 groups, so you need to disregard their order, which means divide by the number of ways these 3 groups can be ordered. Which is 3!

Example:

A B C D E F G H I

CDE | BGI | AFH

This is 1 option. because order does not matter, we need to eliminate situations such as:

BGI | CDE | AFH

This is the same as counting the number of ways 3 things can be arranged. Permutation of 3 things = 3!.

I believe the answer for 3 groups of 3 out of 9 people, and order does not matter would be 280.

C_9^3 = \frac{9*8*7}{3*2*1} * C_6^3 = \frac{6*5*4}{3*2*1} * C_3^3 = 1

This all comes to be
\frac{9*8*7*6*5*4*1}{3*2*1*3*2*1*3*2*1}

8*7*5

7*40

280
_________________

------------------------------------
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a.

Get the best GMAT Prep Resources with GMAT Club Premium Membership

SVP
Joined: 07 Nov 2007
Posts: 1829
Location: New York
Followers: 25

Kudos [?]: 426 [0], given: 5

Re: gmat club [#permalink]  26 Aug 2008, 11:08
jallenmorris wrote:
arjtryarjtry wrote:
thanks bhushan.. it helped me.
so we have to divide b the no. of grps thus formed, and multiplication should be done for the same no. of grps is it ?
eg
three grps -> 6c2*4c2*2c2/3!
two grps-> 6c3*3c3/3!

suppose it was 1 group of 4 people
is it just 6c4?

Yes, if it is C_6^4 for selecting 1 group of 4 people from 6. The thing that I forgot with regard to this question was the significance of "order does not matter." I was thinking that C_6^3 was correct, but didn't remember to divide by the permutations of the order of the group. For instance.

3 groups of 3 from 9 people

\frac{C_3^9 * C_3^6 * C_3^3}{3!}

The first group formed can have any 3 of the 9. next group will be any 3 of the 6 remaining, and the last group is formed by selection the 3 from the 6, because you have 3 remaining, which is why it will always be 1. But then you have 3 groups, so you need to disregard their order, which means divide by the number of ways these 3 groups can be ordered. Which is 3!

Example:

A B C D E F G H I

CDE | BGI | AFH

This is 1 option. because order does not matter, we need to eliminate situations such as:

BGI | CDE | AFH

This is the same as counting the number of ways 3 things can be arranged. Permutation of 3 things = 3!.

I believe the answer for 3 groups of 3 out of 9 people, and order does not matter would be 280.

C_9^3 = \frac{9*8*7}{3*2*1} * C_6^3 = \frac{6*5*4}{3*2*1} * C_3^3 = 1

This all comes to be
\frac{9*8*7*6*5*4*1}{3*2*1*3*2*1*3*2*1}

8*7*5

7*40

280

good job Allen and Bhusan.

What if question says
How many ways are there to split a group of 6 boys into two groups of 4 boys in one group and 2 boys in another group? (The order of the groups does not matter)
_________________

Your attitude determines your altitude
Smiling wins more friends than frowning

Manager
Joined: 15 Jul 2008
Posts: 210
Followers: 3

Kudos [?]: 21 [0], given: 0

Re: gmat club [#permalink]  26 Aug 2008, 12:00
x2suresh wrote:

good job Allen and Bhusan.

What if question says
How many ways are there to split a group of 6 boys into two groups of 4 boys in one group and 2 boys in another group? (The order of the groups does not matter)

Actually I should not have used the word "order" in the earlier posts. Multiple counting would have been a better choice of word. Double counting when there are two groups, 3! counting when there are 3 groups..... so on and so forth.

Now coming to your question.. two groups - one with 4 ppl and the other with 2 ppl.
In this case the answer will be 6C4*2C2. Here there is no scope of double counting of groups as the two groups are of unequal number. I am not sure if this is clear. But I cant think of any other way except using exaples of the two cases... which is a last resort.
SVP
Joined: 07 Nov 2007
Posts: 1829
Location: New York
Followers: 25

Kudos [?]: 426 [0], given: 5

Re: gmat club [#permalink]  26 Aug 2008, 12:04
bhushangiri wrote:
x2suresh wrote:

good job Allen and Bhusan.

What if question says
How many ways are there to split a group of 6 boys into two groups of 4 boys in one group and 2 boys in another group? (The order of the groups does not matter)

Actually I should not have used the word "order" in the earlier posts. Multiple counting would have been a better choice of word. Double counting when there are two groups, 3! counting when there are 3 groups..... so on and so forth.

Now coming to your question.. two groups - one with 4 ppl and the other with 2 ppl.
In this case the answer will be 6C4*2C2. Here there is no scope of double counting of groups as the two groups are of unequal number. I am not sure if this is clear. But I cant think of any other way except using exaples of the two cases... which is a last resort.

agreed. Only when we divide the group into equal number groups....
_________________

Your attitude determines your altitude
Smiling wins more friends than frowning

Re: gmat club   [#permalink] 26 Aug 2008, 12:04
Similar topics Replies Last post
Similar
Topics:
How many ways are there to split a group of 6 boys into two 5 25 Nov 2010, 16:06
2 How many ways are there to split a group of 6 boys into two 6 27 Sep 2010, 13:15
How many ways are there to split 8 boys into 4 groups of 3 1 01 Nov 2007, 06:46
How many ways are there to split a group of 6 boys into two 10 02 Jul 2007, 14:16
In how many ways 5 boys and 6 girls can be seated on 12 5 22 Jan 2006, 00:44
Display posts from previous: Sort by

How many ways are there to split a group of 6 boys into two

 Question banks Downloads My Bookmarks Reviews Important topics

 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®.