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Re: 9 people and Combinatorics [#permalink]
26 Sep 2010, 08:56
3
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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
This post received 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 answer is A
Re: In how many different ways can a group of 9 people be divide [#permalink]
24 Jan 2015, 12:48
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Re: In how many different ways can a group of 9 people be divide [#permalink]
25 Jul 2015, 08:11
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.
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