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Re: 9 people and Combinatorics [#permalink]
26 Sep 2010, 08:56

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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.

Re: 9 people and Combinatorics [#permalink]
15 Jun 2011, 19:06

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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!. _________________

Re: 9 people and Combinatorics [#permalink]
19 Jun 2011, 17:53

Expert's post

1

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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. _________________

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?/

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!) _________________

Re: In how many different ways can a group of 9 people be divide [#permalink]
28 Oct 2013, 11:44

1

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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 answer is A

gmatclubot

Re: In how many different ways can a group of 9 people be divide
[#permalink]
28 Oct 2013, 11:44

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