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In how many different ways can a group of 9 people be divide [#permalink]
 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
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Re: 9 people and Combinatorics [#permalink]
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. Answer: A.
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Re: 9 people and Combinatorics [#permalink]
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. Answer is (A) or 280
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Re: 9 people and Combinatorics [#permalink]
03 Oct 2010, 04:25
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, 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]
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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Re: 9 people and Combinatorics [#permalink]
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 ? 
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Re: 9 people and Combinatorics [#permalink]
19 Jun 2011, 18:53
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.
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Re: 9 people and Combinatorics [#permalink]
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 Voodoo
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Re: 9 people and Combinatorics [#permalink]
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]
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: 9 people and Combinatorics
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07 Oct 2012, 10:08
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