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# In how many different ways can a group of 9 people be divide

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In how many different ways can a group of 9 people be divide [#permalink]  26 Sep 2010, 08:47
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Question Stats:

40% (01:31) correct 59% (01:11) wrong based on 96 sessions
In how many different ways can a group of 9 people be divided into 3 groups, with each group containing 3 people?

A. 280
B. 1,260
C. 1,680
D. 2,520
E. 3,360
[Reveal] Spoiler: OA
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Kudos [?]: 17659 [2] , given: 2234

Re: 9 people and Combinatorics [#permalink]  26 Sep 2010, 08:56
2
KUDOS
Expert's post
hemanthp wrote:
In how many different ways can a group of 9 people be divided into 3 groups, with each group containing 3 people?

280
1,260
1,680
2,520
3,360

GENERAL RULE:
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 important is \frac{(mn)!}{(n!)^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 NOT important is \frac{(mn)!}{(n!)^m*m!}.

BACK TO THE ORIGINAL QUESTION:
In original question I think the order is NOT important, as we won't have group #1, #2 and #3. So we should use second formula, mn=9, m=3 groups n=3 objects (people):
\frac{(mn)!}{(n!)^m*m!}=\frac{9!}{(3!)^3*3!}=280.

This can be done in another way as well: \frac{9C3*6C3*3C3}{3!}=280, we are dividing by 3! as there are 3 groups and order doesn't matter.

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Re: 9 people and Combinatorics [#permalink]  01 Oct 2010, 23:41
hemanthp wrote:
In how many different ways can a group of 9 people be divided into 3 groups, with each group containing 3 people?

280
1,260
1,680
2,520
3,360

To divide 9 persons into 3 groups, when the ordering of groups is not important can be done in \frac{1}{3!} * \frac{9!}{(3!)^3} ways.

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Re: 9 people and Combinatorics [#permalink]  03 Oct 2010, 03:25
1
KUDOS
the number of ways to choose 9 people in 3 groups each having 3 people is

9C3 * 6C3 * 3C3 = 280

another way is = (3*3)!/((3!)^3)*3! = 280
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Re: 9 people and Combinatorics [#permalink]  15 Jun 2011, 05:34
Hi Bunuel
Can u pls explain why u r dividing by 3!
we are not ordering here we are only choosing.
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Kudos [?]: 3623 [3] , given: 144

Re: 9 people and Combinatorics [#permalink]  15 Jun 2011, 19:06
3
KUDOS
Expert's post
toughmat wrote:
Hi Bunuel
Can u pls explain why u r dividing by 3!
we are not ordering here we are only choosing.

We divide by 3! because of exactly what you said: "we are not ordering here we are only choosing."
When you say, "9C3 * 6C3 * 3C3," what you are doing is that you are choosing 3 people of 9 for group 1, 3 people out of the leftover 6 people for group 2 and the rest of the three people for group 3. You have inadvertently marked the 3 groups as distinct. But if we want to just divide them in 3 groups without any distinction of group 1, 2 or 3, we need to divide this by 3!.
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Save $100 on Veritas Prep GMAT Courses And Admissions Consulting Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options. Veritas Prep Reviews Manager Joined: 16 Feb 2011 Posts: 197 Schools: ABCD Followers: 1 Kudos [?]: 17 [0], given: 78 Re: 9 people and Combinatorics [#permalink] 18 Jun 2011, 21:14 How can we compute the # of ways in which 9 objects are divided into groups of 4,3 and 2 ? Can you please help ? Here's what I think: let 9 objects be AAAABBBCC Therefore, combinations = 9!/(4!*3!*2!) Correct ? [Order is important] If order is not important, since we have three groups, # of combinations = 9!/[(4!*3!*2!) * (3!)] Correct ? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 4037 Location: Pune, India Followers: 862 Kudos [?]: 3623 [0], given: 144 Re: 9 people and Combinatorics [#permalink] 19 Jun 2011, 17:53 Expert's post voodoochild wrote: How can we compute the # of ways in which 9 objects are divided into groups of 4,3 and 2 ? Can you please help ? Here's what I think: let 9 objects be AAAABBBCC Therefore, combinations = 9!/(4!*3!*2!) Correct ? [Order is important] If order is not important, since we have three groups, # of combinations = 9!/[(4!*3!*2!) * (3!)] Correct ? Actually, in this case the groups are distinct - a group of 4 people, another of 3 people and another of 2 people. A case in which Mr A is in the four person group is different from the one in which he is in 3 person group. So you will not divide by 3! at the end in second case. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Save$100 on Veritas Prep GMAT Courses And Admissions Consulting
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Re: 9 people and Combinatorics [#permalink]  19 Jun 2011, 18:51
VeritasPrepKarishma wrote:
voodoochild wrote:
How can we compute the # of ways in which 9 objects are divided into groups of 4,3 and 2 ? Can you please help ?

Here's what I think:

let 9 objects be AAAABBBCC
Therefore, combinations = 9!/(4!*3!*2!) Correct ? [Order is important]

If order is not important,

since we have three groups,

# of combinations = 9!/[(4!*3!*2!) * (3!)]

Correct ?

Actually, in this case the groups are distinct - a group of 4 people, another of 3 people and another of 2 people. A case in which Mr A is in the four person group is different from the one in which he is in 3 person group. So you will not divide by 3! at the end in second case.

Thanks Karishma. So, are you saying that the order will not matter ? Essentially, # of combinations = 9!/(4!*3!*2!) irrespective of order?/

Thanks
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Re: 9 people and Combinatorics [#permalink]  21 Jun 2011, 01:29
Expert's post
voodoochild wrote:
VeritasPrepKarishma wrote:
voodoochild wrote:
How can we compute the # of ways in which 9 objects are divided into groups of 4,3 and 2 ? Can you please help ?

Here's what I think:

let 9 objects be AAAABBBCC
Therefore, combinations = 9!/(4!*3!*2!) Correct ? [Order is important]

If order is not important,

since we have three groups,

# of combinations = 9!/[(4!*3!*2!) * (3!)]

Correct ?

Actually, in this case the groups are distinct - a group of 4 people, another of 3 people and another of 2 people. A case in which Mr A is in the four person group is different from the one in which he is in 3 person group. So you will not divide by 3! at the end in second case.

Thanks Karishma. So, are you saying that the order will not matter ? Essentially, # of combinations = 9!/(4!*3!*2!) irrespective of order?/

Thanks
Voodoo

Yes, the groups are distinct so no of combinations is 9!/(4!*3!*2!)
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Manager
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Re: 9 people and Combinatorics [#permalink]  07 Oct 2012, 09:08
geturdream wrote:
the number of ways to choose 9 people in 3 groups each having 3 people is

9C3 * 6C3 * 3C3 = 280

another way is = (3*3)!/((3!)^3)*3! = 280

Hi Geturdream,
How comes A = 9C3 * 6C3 * 3C3 = 280 ?
9C3 = 9!/ (6!*3!) = 9*8*7 / (3*2) = 84
6C3 = 6! / (3!*3!) = 6*5*4 / (3*2) = 20
3C3 = 3!/ (3!*0!) = 1
=> A = 84 * 20 = 1680 ?
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Re: In how many different ways can a group of 9 people be divide [#permalink]  28 Oct 2013, 11:44
1
KUDOS
hemanthp wrote:
In how many different ways can a group of 9 people be divided into 3 groups, with each group containing 3 people?

A. 280
B. 1,260
C. 1,680
D. 2,520
E. 3,360

I used a way found in another topic:

How many ways can 1 person be put with the other 8 in groups of 3? 28
How many ways can following person be put with the remaining 5 in groups of 3? (subract first group total) : 10
How many ways can the final 3 be placed into a group of 3? (subtract last 3) : 1
28 * 10 * 1 = 280
Re: In how many different ways can a group of 9 people be divide   [#permalink] 28 Oct 2013, 11:44
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