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In how many ways can 5 different colored marbles be placed in 3 distin
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14 Oct 2013, 08:53
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In 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: In how many ways can 5 different colored marbles be placed in 3 distin
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15 Oct 2013, 02:19
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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Re: In how many ways can 5 different colored marbles be placed in 3 distin
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20 Sep 2014, 13:22
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: In how many ways can 5 different colored marbles be placed in 3 distin
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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: In how many ways can 5 different colored marbles be placed in 3 distin
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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: In how many ways can 5 different colored marbles be placed in 3 distin
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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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Re: In how many ways can 5 different colored marbles be placed in 3 distin
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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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Re: In how many ways can 5 different colored marbles be placed in 3 distin
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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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Re: In how many ways can 5 different colored marbles be placed in 3 distin
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09 Oct 2018, 23:13
VeritasKarishma wrote: 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. Dear Karishma .. I am bitterly confused with this problem . I doubt the approach or the problem has some defect . Please correct me if I am wrong . All the approaches are considering 3 bags as identical , but in problem they are distinct . My approach to this problem : I am considering 5 different marbles a b c d e and 3 different bags 1 2 3 . So first distribute 3 balls one ball each to each bag . select 3 balls to distribut 5C3 and then distribute to 3 different bags in 3! ways . so total 60 ways . Now we are left with 2 balls and 3 bags each containing one ball each . for first ball we have 3 options and second ball again 3 options . 3*3 ways . TOTAL = 60*3*3=540 ways . Please somebody help me in this . This problem just made me mad ..



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Re: In how many ways can 5 different colored marbles be placed in 3 distin
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10 Oct 2018, 00:22
karnaidu wrote: Dear Karishma .. I am bitterly confused with this problem . I doubt the approach or the problem has some defect . Please correct me if I am wrong . All the approaches are considering 3 bags as identical , but in problem they are distinct . My approach to this problem : I am considering 5 different marbles a b c d e and 3 different bags 1 2 3 . So first distribute 3 balls one ball each to each bag . select 3 balls to distribut 5C3 and then distribute to 3 different bags in 3! ways . so total 60 ways . Now we are left with 2 balls and 3 bags each containing one ball each . for first ball we have 3 options and second ball again 3 options . 3*3 ways . TOTAL = 60*3*3=540 ways . Please somebody help me in this . This problem just made me mad .. You have some double counting here. When you select some from a group and distribute and then distribute the rest to the same bags/pockets, there is double counting. Say you distributed a, b, c first such that Bag1 had a Bag2 had b Bag3 had c Then you distributed d and e such that Bag1 got d and bag2 got e. Take another case. Say you distributed d, b and c first such that Bag1 had d Bag2 had b Bag3 had c Then you distributed a and e such that Bag1 got a and bag2 got e. Note that both cases have exactly the same end result. But you would count them as two separate cases. Similarly, there will be other cases which will be double counted. Hence this method will not work.
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Re: In how many ways can 5 different colored marbles be placed in 3 distin
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10 Oct 2018, 00:53
sxyz wrote: 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... Hii .. In this approach you are missing arrangement of bags . You have to take care of that too ..



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Re: In how many ways can 5 different colored marbles be placed in 3 distin
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18 Oct 2018, 11:16
Hi. I don't understand why this has been multiplied by 3!/2! for 221 combination. Plus, shouldn't 5c2*3c2 be multiplied by 2! as the pockets are distinct, and the arrangement would matter? Thanks



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Re: In how many ways can 5 different colored marbles be placed in 3 distin
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05 May 2019, 05:41
sxyz wrote: 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... Hi, I have a doubt. Here, the case {3,1,1} = \(5C3*2C1*1C1 = 10*2*1 = 20\) And this is multiplied by 3 (coz three different pockets.) But generally, in case of distribution between identical pockets/bags/groups etc., the value is divided by the similar ones. Such as, in here, {3,1,1} has 2 similarly filled pockets, and hence, divided by 2!. Is it not being done here like that because the pockets are all different here?
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Re: In how many ways can 5 different colored marbles be placed in 3 distin
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05 May 2019, 13:48
Hi,
Can we solve it the following way?
Total number of possible combinations: \(3^{5}=243\) Combinations that do not satisfy given condition (at least 1 marble in each pocket): \(410\), \(320\) and \(500\). 1) 410. 4 marbles can be allocated to 3 different pockets, the rest  1  can also be allocated to 3 different pockets. Thus, number of unsatisfactory combinations in for this case = \(3*3*5 = 45\)  multiply by 5, since marbles can also differ between the pockets. 2) 320. The same holds true here = \(3*3*5 = 45\). 3) 500. This scenario is easy = \(3\) different cases are possible.
Thus, \(24345453=150\).



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Re: In how many ways can 5 different colored marbles be placed in 3 distin
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26 Aug 2019, 13:19
praffulpatel wrote: In 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 given: 5 dif marbs 3 dif pockets with at least 1 marble each \([3,1,1]…5c3•2c1•1=10•2=20…•arrangements[3,1,1]=20•(3!/2!)=60\) (3!=num.pockets; 2!=num.duplicates [1,1]) \([2,2,1]…5c2•3c2•1=10•3=30…•arrangements[2,2,1]=30•(3!/2!)=90\) (3!=num.pockets; 2!=num.duplicates [2,2]) \(total.arrangements=60+90=150\) Answer (D)



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Re: In how many ways can 5 different colored marbles be placed in 3 distin
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03 Oct 2019, 10:28
We have two cases: The first case has one pocket with 3 marbles, 1 marble in each of the two remaining pockets. The second case has two pockets with 2 marbles each, and one pocket with one marble. We always add cases together to obtain the total number of possibilities.
Case 1: (3,1,1) First choose which of the 3 pockets will get three marbles. There are 3C1 =3 ways to do this. Then, choose the number of ways we can place 3 marbles into that box from 5 marbles. There are 5C3 = 10 ways to do this. Next we must place a marble in the next to the last box. There are 2C1 = 2 ways to do this. Then there is 1C1 = 1 way to place the last marble into the last pocket. Each time we make a subsequent selection when working through a case we multiply, so there are 3*10*2*1= 60 ways to place the marbles in case 1
Case 2: (2,2,1) First we must select which two pockets get two marbles each. There are 3C2 = 3 ways to do this. Then, we must choose two marbles to go in the first of two pockets. There are 5C2=10 ways to do this. From the 3 remaining marbles, there are 3C2 = 3 ways to place two marbles in the second pocket. Finally there is 1C1 =1 way to place the last marble in the last box. Multiplying, we have 3*10*3*1 = 90 ways to place the marbles in case two.
Thus there are 60+90 = 150 ways we can place the marbles



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Re: In how many ways can 5 different colored marbles be placed in 3 distin
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04 Oct 2019, 22:44
praffulpatel wrote: In 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 Solved it a little differently than others from what I read . So will share. I first made sure that each of the pocket has at least 1 marble. That is 5C3 * 3! Then you can either distribute the remaining 2 as 2, 0, 0 or 1, 1, 0 That is 3C1 or 3C2 * 2 ! (3C1 is the no. of ways of choosing a pocket to put everything in. ) That gives 150. D !



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Re: In how many ways can 5 different colored marbles be placed in 3 distin
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11 Oct 2019, 08:56
P1 can have either 1 marble or 2 marble or 3 marble.
Therefore,
P1 can have 1 marble in 5 ways as all are of different colors.
Or
P1 can have 2 marbles in 5C2 ways (not 5P2 as order is not important; meaning P1 can have m1m2 or m2m1 and it will mean the same thing). 10 ways.
Or
P1 can have 3 marbles in 5C3 ways. 10 ways.
Hence P1 can have marbles in 25 ways.
Every time we select these marbles their arrangement is also important among the 3 pockets. Because if P1 selects M1 first then it will have repercussion on the selection by the next 2 pockets. Similarly if P1 selects M2 first it will again have an effect on the selection by the next 2 pockets. This means that the arrangement among the 3 pockets shall be considered.
So now we will arrange these 25 number of ways among the 3 pockets.
Therefore,
25 x 3! = 25 X 6 = 150.
Kindly correct if there is anything wrong with my logic. Thanks.
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