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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]

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26 Sep 2010, 09:47
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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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Re: 9 people and Combinatorics [#permalink]

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26 Sep 2010, 09:56
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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]

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02 Oct 2010, 00:41
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]

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03 Oct 2010, 04:25
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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]

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15 Jun 2011, 06:34
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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Re: 9 people and Combinatorics [#permalink]

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15 Jun 2011, 20:06
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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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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 16 Feb 2011 Posts: 193 Schools: ABCD Followers: 1 Kudos [?]: 162 [0], given: 78 Re: 9 people and Combinatorics [#permalink] ### Show Tags 18 Jun 2011, 22:14 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: 6966 Location: Pune, India Followers: 2027 Kudos [?]: 12741 [0], given: 221 Re: 9 people and Combinatorics [#permalink] ### Show Tags 19 Jun 2011, 18:53 Expert's post 1 This post was BOOKMARKED 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 Get started with Veritas Prep GMAT On Demand for$199

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Re: 9 people and Combinatorics [#permalink]

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19 Jun 2011, 19:51
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]

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21 Jun 2011, 02:29
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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Re: 9 people and Combinatorics [#permalink]

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07 Oct 2012, 10:08
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]

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28 Oct 2013, 12:44
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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
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Re: In how many different ways can a group of 9 people be divide [#permalink]

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24 Jan 2015, 13:48
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Re: In how many different ways can a group of 9 people be divide [#permalink]

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25 Jul 2015, 09:11
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gamelord wrote:
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 ?

1680 needs to be divided by 3! because the order of the 3 groups do not matter for this problem. The order of the groups, i.e. G1G2G3 vs G2G3G1 (there are 4 more possible combinations) do not matter.

$$\frac{9C3 * 6C3 * 3C3}{3!} = \frac{84*20*1}{3!} = \frac{1680}{3!} = 280$$
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Re: In how many different ways can a group of 9 people be divide [#permalink]

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31 Jul 2016, 00:08
Hello from the GMAT Club BumpBot!

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Re: In how many different ways can a group of 9 people be divide   [#permalink] 31 Jul 2016, 00:08
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