Your answer of $7^5$ is correct since there are seven choices for each of the five marbles.
Line up the marbles in some order. Since the marbles may not share a box, we have seven choices for the first marble, six choices for the second marble, five choices for the third marble, four choices for the fourth marble, and three choices for the fifth marble. Hence, the number of ways the marbles can be distributed is $$7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 = \frac{7!}{2!} = \frac{7!}{(7 - 5)!} = P(7, 5)$$ not $$P(7, 2) = \frac{7!}{(7 - 2)!} = \frac{7!}{5!} = 7 \cdot 6$$
We must select five of the seven boxes in which to place the marbles, which can be done in
ways.
Let $x_k$, $1 \leq k \leq 7$, be the number of marbles placed in the $k$th box. Then $$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 = 5$$ which is an equation in the nonnegative integers. A particular solution corresponds to the placement of six addition signs in a row of seven ones. For instance, $$1 + + 1 1 1 + 1 + + +$$ corresponds to $x_1 = 1$, $x_2 = 0$, $x_3 = 3$, $x_4 = 1$, $x_5 = x_6 = x_7 = 0$, while $$+ 1 + 1 + + 1 + 1 + 1$$ corresponds to $x_1 = 0$, $x_2 = x_3 = 1$, $x_4 = 0$, $x_5 = x_6 = x_7 = 1$. Thus, the number of ways we can distribute the marbles is the number of ways we can insert six addition signs in a row of five ones, which is
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In how many ways can 5 different colored marbles be placed in 3 distin [#permalink]
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Question Stats: Hide Show timer StatisticsIn 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 [#permalink]
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 casesCase-1: One Pocket with 3 marbles and two pockets with 1 marble each No. of Arrangements = 5C3 * 3C1 * 2! = 10*3*2 = 605C3 - No. of ways of choosing 3 out of 5 marbles which have to go in one pocket3C1 - No. of ways of choosing 1 out of 3 pockets in which 3 marbles have to go2! - No. of ways of arranging remaining 2 marbles between remaining two pockets which get one marble eachCase-2: Two Pockets with 2 marbles each and one pockets with 1 marble No. of Arrangements = 5C2 * 3C2 * 3C2 = 10*3*3 = 905C2 - No. of ways of choosing 2 out of 5 marbles which have to go in one pocket3C2 - No. of ways of choosing 2 out of remaining 3 marbles which have to go in second pocket3C2 - No. of ways of choosing 2 out of 3 pockets in each of which 2 marbles have to goTotal cases = 60+90 = 150Answer: option D _________________
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Re: In how many ways can 5 different colored marbles be placed in 3 distin [#permalink] How can we fill three pockets with 5 marbles?3 + 1 + 12 + 2 + 1With 3,1,1 distribution:# of ways to select 3 from 55!/3!2! = 10# of ways to select 1 ball from 22!/1! = 2# of ways to select 1 ball from 11!/1! = 1How many ways to distribute 3,1 and 1 to 3 boxes? 3!/2! = 310*2*3 = 60With 2,2,1 distribution:How many ways to select 2 from 5?5!/2!3! = 10# of ways to select 2 from 33!/2!1! = 3# of ways to select 1 from 11How many ways to distribute 2,2,1 to 3 boxes? 3!/2! = 310*3*3=9090+60 = 150 Answer: 150. D
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Re: In how many ways can 5 different colored marbles be placed in 3 distin [#permalink] This is a tricky question!
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Re: In how many ways can 5 different colored marbles be placed in 3 distin [#permalink] GMATinsight, Bunuel, VeritasPrepKarishma, I'm hoping one of you can explain why the 1-1-1 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 1-1-1 combination since a marble arrangement of Green-Blue-Red in pockets one-two-three is different from Blue-Red-Green in pockets one-two-three and so forth. For the 1-1-1 combination, we will have: 5!/(5-3)! = 60 different arrangements.Then we come to the 3-1-1 and 2-2-1 combinations. For the 3-1-1 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 one-two-three. 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 3-1-1 combination, we have: 5C3 * 2C1 * 1C1 * 3!/2! = 60 different arrangements. Similarly, for the 2-2-1 combination we have: 5C2 * 3C2 * 1C1 * 3!/2! = 90 different arrangements. Finally, we will get 60+60+90 = 210 arrangements if we consider 1-1-1 to be a valid option - since the question does not explicitly exclude this possibility. Could you please explain why we ignored 1-1-1 combination?
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Re: In how many ways can 5 different colored marbles be placed in 3 distin [#permalink]
jhabib wrote: GMATinsight, Bunuel, VeritasPrepKarishma, I'm hoping one of you can explain why the 1-1-1 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 1-1-1 combination since a marble arrangement of Green-Blue-Red in pockets one-two-three is different from Blue-Red-Green in pockets one-two-three and so forth. For the 1-1-1 combination, we will have: 5!/(5-3)! = 60 different arrangements.Then we come to the 3-1-1 and 2-2-1 combinations. For the 3-1-1 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 one-two-three. 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 3-1-1 combination, we have: 5C3 * 2C1 * 1C1 * 3!/2! = 60 different arrangements. Similarly, for the 2-2-1 combination we have: 5C2 * 3C2 * 1C1 * 3!/2! = 90 different arrangements. Finally, we will get 60+60+90 = 210 arrangements if we consider 1-1-1 to be a valid option - since the question does not explicitly exclude this possibility. Could you please explain why we ignored 1-1-1 combination? Hey jhabib,You have to place all the marbles. If you assume the 1-1-1 combination, note that 2 marbles are leftover. But you HAVE TO distribute 5 marbles. So 3-1-1 and 2-2-1 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 [#permalink] Hi jhabib The Question has clearly specified that "ALL the Marbles have to be assigned to 3 pockets" so 1-1-1 needs to be ignored Along with 1-1-1, 1-1-2, 1-2-1 ans 2-1-1 also need to be ignored. jhabib wrote: GMATinsight, I'm hoping one of you can explain why the 1-1-1 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 1-1-1 combination since a marble arrangement of Green-Blue-Red in pockets one-two-three is different from Blue-Red-Green in pockets one-two-three and so forth. For the 1-1-1 combination, we will have: 5!/(5-3)! = 60 different arrangements.Then we come to the 3-1-1 and 2-2-1 combinations. For the 3-1-1 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 one-two-three. 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 3-1-1 combination, we have: 5C3 * 2C1 * 1C1 * 3!/2! = 60 different arrangements. Similarly, for the 2-2-1 combination we have: 5C2 * 3C2 * 1C1 * 3!/2! = 90 different arrangements. Finally, we will get 60+60+90 = 210 arrangements if we consider 1-1-1 to be a valid option - since the question does not explicitly exclude this possibility. Could you please explain why we ignored 1-1-1 combination? _________________
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Re: In how many ways can 5 different colored marbles be placed in 3 distin [#permalink]
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 hiI have seen a solution to a problem similar to this one elsewhere on the forumlet me explain it to you5 different colored marbles can be placed in 3 distinct pockets without any restriction is= 3^5 = 243as 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 3^5so lets get goingnumber 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 = 902 has been subtracted to eliminate the possibility that any pocket out of 2 can get all the marbles3 has been multiplied with the whole expression, because 2 pockets out of 3 have been selected now we are in businessthe answer is= 243 - 90 - 3= 150 (D)hope this helpsthankscheers, 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 [#permalink]
VeritasKarishma wrote: jhabib wrote: GMATinsight, Bunuel, VeritasPrepKarishma, I'm hoping one of you can explain why the 1-1-1 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 1-1-1 combination since a marble arrangement of Green-Blue-Red in pockets one-two-three is different from Blue-Red-Green in pockets one-two-three and so forth. For the 1-1-1 combination, we will have: 5!/(5-3)! = 60 different arrangements.Then we come to the 3-1-1 and 2-2-1 combinations. For the 3-1-1 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 one-two-three. 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 3-1-1 combination, we have: 5C3 * 2C1 * 1C1 * 3!/2! = 60 different arrangements. Similarly, for the 2-2-1 combination we have: 5C2 * 3C2 * 1C1 * 3!/2! = 90 different arrangements. Finally, we will get 60+60+90 = 210 arrangements if we consider 1-1-1 to be a valid option - since the question does not explicitly exclude this possibility. Could you please explain why we ignored 1-1-1 combination? Hey jhabib,You have to place all the marbles. If you assume the 1-1-1 combination, note that 2 marbles are leftover. But you HAVE TO distribute 5 marbles. So 3-1-1 and 2-2-1 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 [#permalink]
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 ..
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Re: In how many ways can 5 different colored marbles be placed in 3 distin [#permalink]
sxyz wrote: This is a tricky question! 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 [#permalink] Hi. I don't understand why this has been multiplied by 3!/2! for 2-2-1 combination. Plus, shouldn't 5c2*3c2 be multiplied by 2! as the pockets are distinct, and the arrangement would matter? Thanks
Re: In how many ways can 5 different colored marbles be placed in 3 distin [#permalink]
sxyz wrote: This is a tricky question! 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 [#permalink] 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): \(4-1-0\), \(3-2-0\) and \(5-0-0\).1) 4-1-0. 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) 3-2-0. The same holds true here = \(3*3*5 = 45\).3) 5-0-0. This scenario is easy = \(3\) different cases are possible. Thus, \(243-45-45-3=150\).
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Re: In how many ways can 5 different colored marbles be placed in 3 distin [#permalink]
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 [#permalink] 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 1Case 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 [#permalink]
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, 0That 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 [#permalink] 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. OrP1 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. OrP1 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. Posted from my mobile device
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Re: In how many ways can 5 different colored marbles be placed in 3 distin [#permalink]
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 Solution attached
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In how many ways can 5 different colored marbles be placed in 3 distin [#permalink]
VeritasKarishma wrote: jhabib wrote: GMATinsight, Bunuel, VeritasPrepKarishma, I'm hoping one of you can explain why the 1-1-1 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 1-1-1 combination since a marble arrangement of Green-Blue-Red in pockets one-two-three is different from Blue-Red-Green in pockets one-two-three and so forth. For the 1-1-1 combination, we will have: 5!/(5-3)! = 60 different arrangements.Then we come to the 3-1-1 and 2-2-1 combinations. For the 3-1-1 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 one-two-three. 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 3-1-1 combination, we have: 5C3 * 2C1 * 1C1 * 3!/2! = 60 different arrangements. Similarly, for the 2-2-1 combination we have: 5C2 * 3C2 * 1C1 * 3!/2! = 90 different arrangements. Finally, we will get 60+60+90 = 210 arrangements if we consider 1-1-1 to be a valid option - since the question does not explicitly exclude this possibility. Could you please explain why we ignored 1-1-1 combination? Hey jhabib,You have to place all the marbles. If you assume the 1-1-1 combination, note that 2 marbles are leftover. But you HAVE TO distribute 5 marbles. So 3-1-1 and 2-2-1 are the only possibilities. All 5 marbles must be distributed. Hi VeritasKarishma Why do we multiply the selection by 3 and not 3!in case 1 (3,1,1) no. of ways= 5C3*2C1*3! (5C3 to select 3 out of 5 diff combo of marbles, 2C1 to select 1 marble and 3! to arrange)Pls explain!
In how many ways can 5 different colored marbles be placed in 3 distin [#permalink] 03 May 2020, 03:44 |
In how many ways can 5 different colored marbles be arranged in a row
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