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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
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Re: 9 people and Combinatorics [#permalink]
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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]
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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 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]
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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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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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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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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]
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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 Voodoo



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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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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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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



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Re: In how many different ways can a group of 9 people be divide [#permalink]
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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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19 Jul 2017, 01:16
Bunuel wrote: 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 \(\frac{(mn)!}{(n!)^m * m!}\) I don't think that's right. For example, let's find out in how many ways 4 different items can be split into 2 groups of 2, when the order is NOT important, using the 2nd formula. \(\frac{(mn)!}{(n!)^m*m!}\) = \(\frac{(4)!}{(2!)^2*2!}\) = 3 This result can be disproved very easily graphically: 1. Draw a square 2. Draw an x through the square drawing 2 lines, corner to corner 3. Represent the 4 objects on each corner of the square 4. Number the lines that link all the square's corners together There are 6 lines which represent all 6 possible groups of 2 objects; every object is connected by a line to all other objects a single time, forming 6 possible groups of 2. The answer is 6. Another solution is 4C2 * 2C2 = 6 nCr = \(\frac{(n!)}{(r!(nr)!)}\) = \(\frac{(4!)}{(r!(42)!)}\) * \(\frac{(2!)}{(r!(22)!)}\) = 6 The groups have not been ordered yet and there is no need to divide by anything else; we proved graphically that 6 is the answer so dividing by 2! or anything else would lead to an incorrect result. If the order was important, there would be \(6!/(62)! = 30\) ways to organize 4 objects into 2 groups of 2.[/b] There would be 2 positions to fill with 6 possible groups. 6 possible groups in the 1st position and 5 possible groups in the 2nd position. nPr= \(\frac{n!}{(nr)!}\) = \(\frac{6!}{(62)!}\) = 6 * 5 = 30 The 1st formula actually represents the number of ways in which the order is NOT important\(\frac{(mn)!}{(n!)^m}\) = \(\frac{(4)!}{(2!)^2}\) = 6 THE ORIGINAL QUESTION:The order is not important in this question.nCr = \(\frac{(n!)}{(r!(nr)!)}\) = \(\frac{(9!)}{(3!(93)!)}\) * \(\frac{(6!)}{(3!(63)!)}\) * \(\frac{(3!)}{(3!(33)!)}\) = 1680 or, alternatively: \(\frac{(mn)!}{(n!)^m}\) = \(\frac{(9)!}{(3!)^3}\) = 1680 If the order mattered:nPr= \(\frac{n!}{(nr)!}\) = \(\frac{1680!}{(16803)!}\) = 1680 * 1679 * 1678 = 4733168160



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In how many different ways can a group of 9 people be divide [#permalink]
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19 Jul 2017, 01:34
dontcountonme wrote: Bunuel wrote: 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 \(\frac{(mn)!}{(n!)^m * m!}\) I don't think that's right. For example, let's find out in how many ways 4 different items can be split into 2 groups of 2, when the order is NOT important, using the 2nd formula. \(\frac{(mn)!}{(n!)^m*m!}\) = \(\frac{(4)!}{(2!)^2*2!}\) = 3 This result can be disproved very easily graphically: 1. Draw a square 2. Draw an x through the square drawing 2 lines, corner to corner 3. Represent the 4 objects on each corner of the square 4. Number the lines that link all the square's corners together There are 6 lines which represent all 6 possible groups of 2 objects; every object is connected by a line to all other objects a single time, forming 6 possible groups of 2. The answer is 6. No you are not right. There are two formulas. The first one for which the order of the groups IS important and the second one for which the order of the groups is NOT important. Case 1  the order of the groups IS important: \(\frac{(mn)!}{(n!)^m}\)2*2 = 4 different items can be divided equally into 2 groups and the order of the groups IS important \(\frac{4!}{(2!)^2}=6\). GROUP 1  GROUP 2 {AB}  {CD} {CD}  {AB} {AC}  {BD} {BD}  {AC} {AD}  {BC} {BC}  {AD} Case 2  the order of the groups is NOT important: \(\frac{(mn)!}{(n!)^m * m!}\)2*2 = 4 different items can be divided equally into 2 groups and the order of the groups is NOT important \(\frac{4!}{(2!)^2*2}=3\). {AB}  {CD} {AC}  {BD} {AD}  {BC} P.S. The correct answer is A, not C. Please refer to several different solutions presented above.
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Re: In how many different ways can a group of 9 people be divide [#permalink]
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VeritasPrepKarishma wrote: 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!. Can we apply the same logic to the below question. A menu consisting of 2 types of breakfasts, 3 types of lunch and 4 types of dinner has to be selected from 10 types of breakfast, 11 types of lunch and 12 types of dinner. Will the solution be 10c*11c3*12c4/3! ?



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Re: In how many different ways can a group of 9 people be divide [#permalink]
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Bunuel wrote: dontcountonme wrote: Bunuel wrote: 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 \(\frac{(mn)!}{(n!)^m * m!}\) I don't think that's right. For example, let's find out in how many ways 4 different items can be split into 2 groups of 2, when the order is NOT important, using the 2nd formula. \(\frac{(mn)!}{(n!)^m*m!}\) = \(\frac{(4)!}{(2!)^2*2!}\) = 3 This result can be disproved very easily graphically: 1. Draw a square 2. Draw an x through the square drawing 2 lines, corner to corner 3. Represent the 4 objects on each corner of the square 4. Number the lines that link all the square's corners together There are 6 lines which represent all 6 possible groups of 2 objects; every object is connected by a line to all other objects a single time, forming 6 possible groups of 2. The answer is 6. No you are not right. There are two formulas. The first one for which the order of the groups IS important and the second one for which the order of the groups is NOT important. Case 1  the order of the groups IS important: \(\frac{(mn)!}{(n!)^m}\)2*2 = 4 different items can be divided equally into 2 groups and the order of the groups IS important \(\frac{4!}{(2!)^2}=6\). GROUP 1  GROUP 2 {AB}  {CD} {CD}  {AB} {AC}  {BD} {BD}  {AC} {AD}  {BC} {BC}  {AD} Case 2  the order of the groups is NOT important: \(\frac{(mn)!}{(n!)^m * m!}\)2*2 = 4 different items can be divided equally into 2 groups and the order of the groups is NOT important \(\frac{4!}{(2!)^2*2}=3\). {AB}  {CD} {AC}  {BD} {AD}  {BC} P.S. The correct answer is A, not C. Please refer to several different solutions presented above. Can we apply the same logic to the below question. A menu consisting of 2 types of breakfasts, 3 types of lunch and 4 types of dinner has to be selected from 10 types of breakfast, 11 types of lunch and 12 types of dinner. Will the solution be 10c*11c3*12c4/3! ?




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