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How many ways can 5 different colored marbles be placed in 3 [#permalink]
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14 Oct 2013, 08:53
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How many ways can 5 different colored marbles be placed in 3 distinct pockets such that any pocket contains at least 1 marble? (A) 60 (B) 90 (C) 120 (D) 150 (E) 180
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Re: How many ways can 5 different colored marbles be placed in 3 [#permalink]
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15 Oct 2013, 02:19
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How can we fill three pockets with 5 marbles? 3 + 1 + 1 2 + 2 + 1 With 3,1,1 distribution: # of ways to select 3 from 5 5!/3!2! = 10 # of ways to select 1 ball from 2 2!/1! = 2 # of ways to select 1 ball from 1 1!/1! = 1 How many ways to distribute 3,1 and 1 to 3 boxes? 3!/2! = 3 10*2*3 = 60 With 2,2,1 distribution: How many ways to select 2 from 5? 5!/2!3! = 10 # of ways to select 2 from 3 3!/2!1! = 3 # of ways to select 1 from 1 1 How many ways to distribute 2,2,1 to 3 boxes? 3!/2! = 3 10*3*3=90 90+60 = 150 Answer: 150. D
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How many ways can 5 different colored marbles be placed in 3 [#permalink]
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20 Sep 2014, 13:22
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This is a tricky question! At least for people like me, who are from nonQuant background... I appreciate Igotthis's post but it appeared a little bit complicated to me (may be because I am not smart enough ) Anyway this is how I solved it...
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Re: How many ways can 5 different colored marbles be placed in 3 [#permalink]
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08 Aug 2015, 07:18
praffulpatel wrote: How many ways can 5 different colored marbles be placed in 3 distinct pockets such that any pocket contains at least 1 marble?
(A) 60 (B) 90 (C) 120 (D) 150 (E) 180 We have 5 marbles and 3 pockets So we have two cases Case1: One Pocket with 3 marbles and two pockets with 1 marble eachNo. of Arrangements = 5C3 * 3C1 * 2! = 10*3*2 = 60 5C3  No. of ways of choosing 3 out of 5 marbles which have to go in one pocket 3C1  No. of ways of choosing 1 out of 3 pockets in which 3 marbles have to go 2!  No. of ways of arranging remaining 2 marbles between remaining two pockets which get one marble each Case2: Two Pockets with 2 marbles each and one pockets with 1 marbleNo. of Arrangements = 5C2 * 3C2 * 3C2 = 10*3*3 = 90 5C2  No. of ways of choosing 2 out of 5 marbles which have to go in one pocket 3C2  No. of ways of choosing 2 out of remaining 3 marbles which have to go in second pocket 3C2  No. of ways of choosing 2 out of 3 pockets in each of which 2 marbles have to go Total cases = 60+90 = 150 Answer: option D
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Re: How many ways can 5 different colored marbles be placed in 3 [#permalink]
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09 Aug 2015, 14:23
GMATinsight, Bunuel, VeritasPrepKarishma, I'm hoping one of you can explain why the 111 combination is ignored in answering this question. The question asks: Quote: How many ways can 5 different colored marbles be placed in 3 distinct pockets such that any pocket contains at least 1 marble? Would the answer not include 5P3? Since we can have at least 1 marble (i.e. exactly one marble) in each spot and the question does not specify that we must use all five of the marbles. Additionally, I think that permutation is the right way of counting for the 111 combination since a marble arrangement of GreenBlueRed in pockets onetwothree is different from BlueRedGreen in pockets onetwothree and so forth. For the 111 combination, we will have: 5!/(53)! = 60 different arrangements. Then we come to the 311 and 221 combinations. For the 311 combination, in pocket one we can have any three of the five distinct marbles. The order inside one pocket doesn't matter. Therefore, we will use 5C3. For the 2nd pocket, we have two marbles left and we can only pick one because we must leave one for the third pocket. Therefore, we will use 2C1. And for the last (third) pocket we only have one choice. Now, we have 5C3, 2C1 and 1C1 in pockets onetwothree. These can be ordered in 3!/2! different ways because 2C1 = 1C1, and we have already counted the possibility of different colors in each pocket. So for 311 combination, we have: 5C3 * 2C1 * 1C1 * 3!/2! = 60 different arrangements. Similarly, for the 221 combination we have: 5C2 * 3C2 * 1C1 * 3!/2! = 90 different arrangements. Finally, we will get 60+60+90 = 210 arrangements if we consider 111 to be a valid option  since the question does not explicitly exclude this possibility. Could you please explain why we ignored 111 combination?



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Re: How many ways can 5 different colored marbles be placed in 3 [#permalink]
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10 Aug 2015, 00:18
jhabib wrote: GMATinsight, Bunuel, VeritasPrepKarishma, I'm hoping one of you can explain why the 111 combination is ignored in answering this question. The question asks: Quote: How many ways can 5 different colored marbles be placed in 3 distinct pockets such that any pocket contains at least 1 marble? Would the answer not include 5P3? Since we can have at least 1 marble (i.e. exactly one marble) in each spot and the question does not specify that we must use all five of the marbles. Additionally, I think that permutation is the right way of counting for the 111 combination since a marble arrangement of GreenBlueRed in pockets onetwothree is different from BlueRedGreen in pockets onetwothree and so forth. For the 111 combination, we will have: 5!/(53)! = 60 different arrangements. Then we come to the 311 and 221 combinations. For the 311 combination, in pocket one we can have any three of the five distinct marbles. The order inside one pocket doesn't matter. Therefore, we will use 5C3. For the 2nd pocket, we have two marbles left and we can only pick one because we must leave one for the third pocket. Therefore, we will use 2C1. And for the last (third) pocket we only have one choice. Now, we have 5C3, 2C1 and 1C1 in pockets onetwothree. These can be ordered in 3!/2! different ways because 2C1 = 1C1, and we have already counted the possibility of different colors in each pocket. So for 311 combination, we have: 5C3 * 2C1 * 1C1 * 3!/2! = 60 different arrangements. Similarly, for the 221 combination we have: 5C2 * 3C2 * 1C1 * 3!/2! = 90 different arrangements. Finally, we will get 60+60+90 = 210 arrangements if we consider 111 to be a valid option  since the question does not explicitly exclude this possibility. Could you please explain why we ignored 111 combination? Hey jhabib, You have to place all the marbles. If you assume the 111 combination, note that 2 marbles are leftover. But you HAVE TO distribute 5 marbles. So 311 and 221 are the only possibilities. All 5 marbles must be distributed.
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How many ways can 5 different colored marbles be placed in 3 [#permalink]
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10 Aug 2015, 04:21
Hi jhabibThe Question has clearly specified that "ALL the Marbles have to be assigned to 3 pockets" so 111 needs to be ignored Along with 111, 112, 121 ans 211 also need to be ignored. I hope it helps! jhabib wrote: GMATinsight, I'm hoping one of you can explain why the 111 combination is ignored in answering this question. The question asks: Quote: How many ways can 5 different colored marbles be placed in 3 distinct pockets such that any pocket contains at least 1 marble? Would the answer not include 5P3? Since we can have at least 1 marble (i.e. exactly one marble) in each spot and the question does not specify that we must use all five of the marbles. Additionally, I think that permutation is the right way of counting for the 111 combination since a marble arrangement of GreenBlueRed in pockets onetwothree is different from BlueRedGreen in pockets onetwothree and so forth. For the 111 combination, we will have: 5!/(53)! = 60 different arrangements. Then we come to the 311 and 221 combinations. For the 311 combination, in pocket one we can have any three of the five distinct marbles. The order inside one pocket doesn't matter. Therefore, we will use 5C3. For the 2nd pocket, we have two marbles left and we can only pick one because we must leave one for the third pocket. Therefore, we will use 2C1. And for the last (third) pocket we only have one choice. Now, we have 5C3, 2C1 and 1C1 in pockets onetwothree. These can be ordered in 3!/2! different ways because 2C1 = 1C1, and we have already counted the possibility of different colors in each pocket. So for 311 combination, we have: 5C3 * 2C1 * 1C1 * 3!/2! = 60 different arrangements. Similarly, for the 221 combination we have: 5C2 * 3C2 * 1C1 * 3!/2! = 90 different arrangements. Finally, we will get 60+60+90 = 210 arrangements if we consider 111 to be a valid option  since the question does not explicitly exclude this possibility. Could you please explain why we ignored 111 combination?
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How many ways can 5 different colored marbles be placed in 3 [#permalink]
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06 Jan 2018, 13:02
praffulpatel wrote: How many ways can 5 different colored marbles be placed in 3 distinct pockets such that any pocket contains at least 1 marble?
(A) 60 (B) 90 (C) 120 (D) 150 (E) 180 hi I have seen a solution to a problem similar to this one elsewhere on the forum let me explain it to you 5 different colored marbles can be placed in 3 distinct pockets without any restriction is = 3^5 = 243 as we are asked to find out the ways in which any pocket must get at least 1 marble, lets find out the ways in which any pocket must not get at least 1 marble, and then subtract the number of ways in which any pocket must not get at least 1 marble from the total number of ways, that is 3^5 so lets get going number of ways in which all marbles can get to 1 pocket is = 3, as there are 3 distinct pocket in total now, number of ways in which 2 pockets can get all the marbles and 1 pocket remains empty is = (2^5  2) * 3 = 90 2 has been subtracted to eliminate the possibility that any pocket out of 2 can get all the marbles 3 has been multiplied with the whole expression, because 2 pockets out of 3 have been selected now we are in business the answer is = 243  90  3 = 150 (D) hope this helps thanks cheers, and do consider some kudos, man




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