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Five balls of different colors are to be placed in three different boxes such that any box contains at least 1 ball . What is the maximum number of different ways in which this can be done? A. 60 B. 90 C. 120 D. 150 E. 180 Please provide a small note of explanation for all the combinations used in the solution.
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avaneeshvyas wrote: Five balls of different colors are to be placed in three different boxes such that any box contains at least 1 ball . What is the maximum number of different ways in which this can be done?
A. 60 B. 90 C. 120 D. 150 E. 180
Please provide a small note of explanation for all the combinations used in the solution. We can have the following two distributions: 1. 311 one box gets three balls and the remaining two boxes get one ball each.The number of ways to distribute the balls for this case is \(3*C^3_5*2=60\), where 3 is the number of ways to choose which box gets 3 balls (we have 3 boxes, thus 3 choices for that), \(C^3_5\) is the number of ways to choose which 3 balls out of 5 will go to that box, and 2 is the number of ways to distribute the remaining 2 balls in the remaining two boxes. 2. 122 one box gets one ball and the remaining two boxes get two balls each.The number of ways to distribute the balls for this case is \(3*5*C^2_4=90\), where 3 is the number of ways to choose which box gets 1 balls (we have 3 boxes, thus 3 choices for that), 5 is the number of ways to choose which ball out of 5 will go to that box, and \(C^2_4\) is the number of ways to chose which 2 balls out of 4 balls left will go to the second box (the remaining 2 balls will naturally go to the third box). Total: 60+90=150. Answer: D. Hope it's clear.
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Re: Five balls of different colors are to be placed in three [#permalink]
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avaneeshvyas wrote: Five balls of different colors are to be placed in three different boxes such that any box contains at least 1 ball . What is the maximum number of different ways in which this can be done?
A. 60 B. 90 C. 120 D. 150 E. 180
Please provide a small note of explanation for all the combinations used in the solution. Since the balls are all of different colors, let's permute them and then decide how many balls we put in each box. For example, arrange in a row the balls, then decide: two balls go into the first box, next two in the second box, and the last ball goes to the third box. Since in each box there must be at least one ball, we have the possibilities of (2,2,1), (2,1,2), (1,2,2) OR (3,1,1), (1,3,1), (1,1,3) balls in the three boxes. For the 2,2,1 type arrangements, we have [5!/(2!2!1!)]*3 = 90 possibilities. Inside a box, it doesn't matter the order of the balls. For the 3,1,1 type arrangements, we have [5!/(3!1!1!)]*3 = 60 possibilities. Total of 90 + 60 = 150 possibilities. Answer D.
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Re: Five balls of different colors are to be placed in three [#permalink]
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Another way: \(N = 3^5  3*2^5 + 3 = 243  96 + 3 = 150\) \(3^5\)  the total number of ways to put 5 different balls into 3 different boxes. Each ball has 3 possible options (boxes) \(3*2^5  3\)  the total number of ways to put 5 different balls into 1 or 2 different boxes out of 3 (to leave at least 1 box empty).
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Re: Five balls of different colors are to be placed in three [#permalink]
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18 Oct 2012, 20:03
EvaJager wrote: avaneeshvyas wrote: Five balls of different colors are to be placed in three different boxes such that any box contains at least 1 ball . What is the maximum number of different ways in which this can be done?
A. 60 B. 90 C. 120 D. 150 E. 180
Please provide a small note of explanation for all the combinations used in the solution. Since the balls are all of different colors, let's permute them and then decide how many balls we put in each box. For example, arrange in a row the balls, then decide: two balls go into the first box, next two in the second box, and the last ball goes to the third box. Since in each box there must be at least one ball, we have the possibilities of (2,2,1), (2,1,2), (1,2,2) OR (3,1,1), (1,3,1), (1,1,3) balls in the three boxes. For the 2,2,1 type arrangements, we have [5!/(2!2!1!)]*3 = 90 possibilities. Inside a box, it doesn't matter the order of the balls. For the 3,1,1 type arrangements, we have [5!/(3!1!1!)]*3 = 60 possibilities. Total of 90 + 60 = 150 possibilities. Answer D. Could you please break down the highlighted part and explain how are you getting those figures.....



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avaneeshvyas wrote: EvaJager wrote: avaneeshvyas wrote: Five balls of different colors are to be placed in three different boxes such that any box contains at least 1 ball . What is the maximum number of different ways in which this can be done?
A. 60 B. 90 C. 120 D. 150 E. 180
Please provide a small note of explanation for all the combinations used in the solution. Since the balls are all of different colors, let's permute them and then decide how many balls we put in each box. For example, arrange in a row the balls, then decide: two balls go into the first box, next two in the second box, and the last ball goes to the third box. Since in each box there must be at least one ball, we have the possibilities of (2,2,1), (2,1,2), (1,2,2) OR (3,1,1), (1,3,1), (1,1,3) balls in the three boxes. For the 2,2,1 type arrangements, we have [5!/(2!2!1!)]*3 = 90 possibilities. Inside a box, it doesn't matter the order of the balls. For the 3,1,1 type arrangements, we have [5!/(3!1!1!)]*3 = 60 possibilities. Total of 90 + 60 = 150 possibilities. Answer D. Could you please break down the highlighted part and explain how are you getting those figures..... If there would not have been boxes, 5 different balls we could arrange in 5! = 120 ways. Now look at a given arrangement of the 5 balls, let's say ABCDE. Because we have three different boxes, then the following arrangements are possible: AB CD E; AB C DE; A BC DE  these are all 2,2,1 type scenarios and they are considered different arrangements (this would give 5! * 3 possibilities), EXCEPT THAT, inside each box, if we have more than one ball, order doesn't matter. So, when in a box we have for example AB, this is the same as BA, we don't mind whether A or B was put first into the box. Therefore, we should divide by k! whenever we have in a box k balls. 1! doesn't change the value of the expression, I left them just for the completeness of the formula. Similar reasoning applies for the 3,1,1 scenario.
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Re: Five balls of different colors are to be placed in three [#permalink]
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19 Oct 2012, 19:21
EvaJager wrote: avaneeshvyas wrote: Five balls of different colors are to be placed in three different boxes such that any box contains at least 1 ball . What is the maximum number of different ways in which this can be done?
A. 60 B. 90 C. 120 D. 150 E. 180
Please provide a small note of explanation for all the combinations used in the solution. Since the balls are all of different colors, let's permute them and then decide how many balls we put in each box. For example, arrange in a row the balls, then decide: two balls go into the first box, next two in the second box, and the last ball goes to the third box. Since in each box there must be at least one ball, we have the possibilities of (2,2,1), (2,1,2), (1,2,2) OR (3,1,1), (1,3,1), (1,1,3) balls in the three boxes. For the 2,2,1 type arrangements, we have [5!/(2!2!1!)]*3 = 90 possibilities. Inside a box, it doesn't matter the order of the balls. For the 3,1,1 type arrangements, we have [5!/(3!1!1!)]*3 = 60 possibilities. Total of 90 + 60 = 150 possibilities. Answer D. Eva, I was able to compute the combinations for distributing balls in 221 and 311 way. However, I used 3! for the arrangement of the boxes. Let's say that the boxes are labeled X Y and Z  and we are using (2,2,1) type of placement of balls. Wouldn't the order of boxes matter? i.e. 3!? That is, I could have any of the three boxes in the first spot, 2 for the second spot and 1 for the third one? Example : abcde are balls and three boxes are xyz => X=[AB] Y=[cd] Z=[e] X[AB] Z[e] Y[cd] .... (repeat two rows three times) = 6 I am a bit confused. Still thinking....Please help me. thanks



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voodoochild wrote: EvaJager wrote: avaneeshvyas wrote: Five balls of different colors are to be placed in three different boxes such that any box contains at least 1 ball . What is the maximum number of different ways in which this can be done?
A. 60 B. 90 C. 120 D. 150 E. 180
Please provide a small note of explanation for all the combinations used in the solution. Since the balls are all of different colors, let's permute them and then decide how many balls we put in each box. For example, arrange in a row the balls, then decide: two balls go into the first box, next two in the second box, and the last ball goes to the third box. Since in each box there must be at least one ball, we have the possibilities of (2,2,1), (2,1,2), (1,2,2) OR (3,1,1), (1,3,1), (1,1,3) balls in the three boxes. For the 2,2,1 type arrangements, we have [5!/(2!2!1!)]*3 = 90 possibilities. Inside a box, it doesn't matter the order of the balls. For the 3,1,1 type arrangements, we have [5!/(3!1!1!)]*3 = 60 possibilities. Total of 90 + 60 = 150 possibilities. Answer D. Eva, I was able to compute the combinations for distributing balls in 221 and 311 way. However, I used 3! for the arrangement of the boxes. Let's say that the boxes are labeled X Y and Z  and we are using (2,2,1) type of placement of balls. Wouldn't the order of boxes matter? i.e. 3!? That is, I could have any of the three boxes in the first spot, 2 for the second spot and 1 for the third one? Example : abcde are balls and three boxes are xyz => X=[AB] Y=[cd] Z=[e] X[AB] Z[e] Y[cd] .... (repeat two rows three times) = 6 I am a bit confused. Still thinking....Please help me. thanks You keep the boxes fixed, you permute the balls. For example, if you have in the boxes AB CD E. This is obtained from ABCDE. You also have CD E AB, which is obtained from CDEAB, an arrangement already included in the 5! permutations of the balls. For any permutation of the balls, in the 2,2,1, scenario (meaning there are two boxes with two balls and one box with one ball), you need the factor of 3 because there are 3 possibilities: AB CD E AB C DE A BC DE So, you have a total of 5! * 3 arrangements. But here we count each arrangement 4 times, which is 2! * 2!. For example: AB in the first box is the same as BA, and CD in the second box is the same as DC. All the arrangements AB CD E BA CD E AB DC E BA DC E should be counted just once. Therefore, 5!*3/(2! * 2! * 1!).
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Re: Five balls of different colors are to be placed in three [#permalink]
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20 Oct 2012, 05:13
avaneeshvyas wrote: Five balls of different colors are to be placed in three different boxes such that any box contains at least 1 ball . What is the maximum number of different ways in which this can be done?
A. 60 B. 90 C. 120 D. 150 E. 180
Please provide a small note of explanation for all the combinations used in the solution. I did it the way walker did since we can avoid enumerating by using that method i.e. we avoid (3, 1, 1), (2, 2, 1) etc. When required to enumerate, there is a chance that we might forget one of the possible combinations (though in this question it doesn't matter much since number of balls is small). Do visit that solution properly too. In this question, consider it an exercise in lots of PnC concepts. In how many ways can you distribute 5 different balls in 3 different boxes such that a box may get no ball? (we will later subtract those cases in which one or two boxes are empty) Each ball can be placed in any of the 3 boxes so there are 3 ways to place a ball. Since there are 5 balls, we can place all the balls in 3*3*3*3*3 = 3^5 ways Only one box empty: In how many ways can you distribute the 5 different balls in 2 different boxes such that both boxes have at least one ball? (we want to find the cases in which one and only one box will be empty) Select the empty box in 3 ways. Each ball can be placed in any of the 2 boxes. So 5 balls can be placed in 2*2*2*2*2 = 2^5 ways. Out of 2^5, subtract those cases where the balls are all in 1 box only. This happens in 2 ways since you can select an empty box again out of the two in 2 ways. Total number of ways = 3*(2^5  2). 2 boxes empty: In how many ways can you distribute the 5 balls in only 1 box? In 3 ways since you select a box in 3 ways. You can distribute 5 balls in 3 boxes such that no box is empty in 3^5  3*(2^5  2)  3 = 243  90  3 = 150 ways For a discussion on distribution of objects, check out these posts too: http://www.veritasprep.com/blog/2011/12 ... 93part1/http://www.veritasprep.com/blog/2011/12 ... spartii/
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Re: Five balls of different colors are to be placed in three [#permalink]
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avaneeshvyas wrote: Five balls of different colors are to be placed in three different boxes such that any box contains at least 1 ball . What is the maximum number of different ways in which this can be done?
A. 60 B. 90 C. 120 D. 150 E. 180
Please provide a small note of explanation for all the combinations used in the solution. This is the beauty of combinatorics: doesn't matter how you execute your plan (here, that of distributing some balls in boxes), if the stages are correctly translated into mathematical expressions, you always get the correct answer. We have already seen three different approaches to solve the question. Just for the fun, I tried to reach the correct answer with another scenario: Two stages: 1) Distribute one ball in each box, to be sure that there is at least one ball in each box. This can be done in 5*4*3 = 60 ways, boxes are distinct, order matters. Continue with stage: 2) Place the remaining 2 balls. Here we have two possibilities: 2a) Either both balls will be placed in one of the 3 boxes  OR  2b) Each ball will be placed in a separate box. For 1 and 2a)  60*3/3 = 60; 3 possibilities to choose the box that will get the remaining 2 balls; divide by 3, because in the box with the 3 balls, it doesn't matter which ball was placed in at stage 1, it could be any of the final 3 balls. For 1 and 2b)  60*3*2/4 = 90; 3 possibilities to choose the box that will not get an extra ball, which is equivalent to choosing 2 boxes that will each get an extra ball; 2 possibilities to place the remaining balls in the 2 different boxes; divide by 4 = 2*2, because in those 2 boxes with 2 balls, it doesn't matter in which order the 2 balls arrived, stage 1 or stage 2b). Again, we obtain the total of 60 + 90 = 150 possibilities.
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Re: Five balls of different colors are to be placed in three [#permalink]
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20 Oct 2012, 19:19
EvaJager wrote: voodoochild wrote: EvaJager wrote: Since the balls are all of different colors, let's permute them and then decide how many balls we put in each box. For example, arrange in a row the balls, then decide: two balls go into the first box, next two in the second box, and the last ball goes to the third box. Since in each box there must be at least one ball, we have the possibilities of (2,2,1), (2,1,2), (1,2,2) OR (3,1,1), (1,3,1), (1,1,3) balls in the three boxes. For the 2,2,1 type arrangements, we have [5!/(2!2!1!)]*3 = 90 possibilities. Inside a box, it doesn't matter the order of the balls. For the 3,1,1 type arrangements, we have [5!/(3!1!1!)]*3 = 60 possibilities.
Total of 90 + 60 = 150 possibilities.
Answer D. Eva, I was able to compute the combinations for distributing balls in 221 and 311 way. However, I used 3! for the arrangement of the boxes. Let's say that the boxes are labeled X Y and Z  and we are using (2,2,1) type of placement of balls. Wouldn't the order of boxes matter? i.e. 3!? That is, I could have any of the three boxes in the first spot, 2 for the second spot and 1 for the third one? Example : abcde are balls and three boxes are xyz => X=[AB] Y=[cd] Z=[e] X[AB] Z[e] Y[cd] .... (repeat two rows three times) = 6 I am a bit confused. Still thinking....Please help me. thanks You keep the boxes fixed, you permute the balls. For example, if you have in the boxes AB CD E. This is obtained from ABCDE. You also have CD E AB, which is obtained from CDEAB, an arrangement already included in the 5! permutations of the balls. For any permutation of the balls, in the 2,2,1, scenario (meaning there are two boxes with two balls and one box with one ball), you need the factor of 3 because there are 3 possibilities: AB CD E AB C DE A BC DE So, you have a total of 5! * 3 arrangements. But here we count each arrangement 4 times, which is 2! * 2!. For example: AB in the first box is the same as BA, and CD in the second box is the same as DC. All the arrangements AB CD E BA CD E AB DC E BA DC E should be counted just once. Therefore, 5!*3/(2! * 2! * 1!). Eva, Thanks for your help. I think that I see your point. Do you think that if the boxes are placed on a rack in a store (order of the three boxes will matter correct?), then we will have to multiply by 3!? That is, for 221 case, 5C2*3C2*1*(3!) Please let me know your thoughts.... { My approach: Here's how I approached this problem : for 221 case, the combinations will be 5C2*3C2*1C1 (choose 2 balls from 5 balls  then  choose 2 balls from the remaining 3 balls  then  choose 1 ball from 1 ball) Now, since the boxes are labeled differently, we need to permute them. I initially thought of multiplying by 3! because of these combinations : X[AB] Y[cd] Z[e] X[AB] Z[e] Y[cd] X[CD] Y[AB] Z[e] X[CD] Z[e] Y[AB] X[e] Y[AB] Z[CD] X[e] Z[CD] Y[AB] ........ditto for 311 case. } I am a bit confused. Thanks



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Re: Five balls of different colors are to be placed in three [#permalink]
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21 Oct 2012, 04:18
voodoochild wrote: EvaJager wrote: voodoochild wrote: Eva, I was able to compute the combinations for distributing balls in 221 and 311 way. However, I used 3! for the arrangement of the boxes. Let's say that the boxes are labeled X Y and Z  and we are using (2,2,1) type of placement of balls. Wouldn't the order of boxes matter? i.e. 3!? That is, I could have any of the three boxes in the first spot, 2 for the second spot and 1 for the third one?
Example : abcde are balls and three boxes are xyz => X=[AB] Y=[cd] Z=[e] X[AB] Z[e] Y[cd] .... (repeat two rows three times) = 6
I am a bit confused. Still thinking....Please help me. thanks
You keep the boxes fixed, you permute the balls. For example, if you have in the boxes AB CD E. This is obtained from ABCDE. You also have CD E AB, which is obtained from CDEAB, an arrangement already included in the 5! permutations of the balls. For any permutation of the balls, in the 2,2,1, scenario (meaning there are two boxes with two balls and one box with one ball), you need the factor of 3 because there are 3 possibilities: AB CD E AB C DE A BC DE So, you have a total of 5! * 3 arrangements. But here we count each arrangement 4 times, which is 2! * 2!. For example: AB in the first box is the same as BA, and CD in the second box is the same as DC. All the arrangements AB CD E BA CD E AB DC E BA DC E should be counted just once. Therefore, 5!*3/(2! * 2! * 1!). Eva, Thanks for your help. I think that I see your point. Do you think that if the boxes are placed on a rack in a store (order of the three boxes will matter correct?), then we will have to multiply by 3!? That is, for 221 case, 5C2*3C2*1*(3!) Please let me know your thoughts.... { My approach: Here's how I approached this problem : for 221 case, the combinations will be 5C2*3C2*1C1 (choose 2 balls from 5 balls  then  choose 2 balls from the remaining 3 balls  then  choose 1 ball from 1 ball) Now, since the boxes are labeled differently, we need to permute them. NO! I initially thought of multiplying by 3! because of these combinations : X[AB] Y[cd] Z[e] X[AB] Z[e] Y[cd] X[CD] Y[AB] Z[e] X[CD] Z[e] Y[AB] X[e] Y[AB] Z[CD] X[e] Z[CD] Y[AB] ........ditto for 311 case. } I am a bit confused. Thanks You don't permute the boxes! Once the boxes are defined, they are unique. You called them X, Y, Z. Place them on a shelf, nailed them down, don't touch them!!! The question is not about arranging the boxes on a shelf, it's about their contents. Even if you would arrange the boxes in a different order, say Y, X, Z, what matters is only their content and not the order in which they are displayed. For the combinatorial approach, see Bunuel and Walker's solutions. Here, I was playing a different game. I was permuting the balls and then decide (define) the boxes. Try to understand each solution separately, and don't mix between them.
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Re: Five balls of different colors are to be placed in three [#permalink]
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21 Oct 2012, 05:22
VeritasPrepKarishma wrote: avaneeshvyas wrote: Five balls of different colors are to be placed in three different boxes such that any box contains at least 1 ball . What is the maximum number of different ways in which this can be done?
A. 60 B. 90 C. 120 D. 150 E. 180
Please provide a small note of explanation for all the combinations used in the solution. I did it the way walker did since we can avoid enumerating by using that method i.e. we avoid (3, 1, 1), (2, 2, 1) etc. When required to enumerate, there is a chance that we might forget one of the possible combinations (though in this question it doesn't matter much since number of balls is small). Do visit that solution properly too. In this question, consider it an exercise in lots of PnC concepts. In how many ways can you distribute 5 different balls in 3 different boxes such that a box may get no ball? (we will later subtract those cases in which one or two boxes are empty) Each ball can be placed in any of the 3 boxes so there are 3 ways to place a ball. Since there are 5 balls, we can place all the balls in 3*3*3*3*3 = 3^5 ways Only one box empty: In how many ways can you distribute the 5 different balls in 2 different boxes such that both boxes have at least one ball? (we want to find the cases in which one and only one box will be empty) Select the empty box in 3 ways. Each ball can be placed in any of the 2 boxes. So 5 balls can be placed in 2*2*2*2*2 = 2^5 ways. Out of 2^5, subtract those cases where the balls are all in 1 box only. This happens in 2 ways since you can select an empty box again out of the two in 2 ways. Total number of ways = 3*(2^5  2). 2 boxes empty: In how many ways can you distribute the 5 balls in only 1 box? In 3 ways since you select a box in 3 ways. You can distribute 5 balls in 3 boxes such that no box is empty in 3^5  3*(2^5  2)  3 = 243  90  3 = 150 ways For a discussion on distribution of objects, check out these posts too: http://www.veritasprep.com/blog/2011/12 ... 93part1/http://www.veritasprep.com/blog/2011/12 ... spartii/Just a remark: ...avoid enumerating by using that method i.e. we avoid (3, 1, 1), (2, 2, 1) etc. When required to enumerate, there is a chance that we might forget one of the possible combinations (though in this question it doesn't matter much since number of balls is small).There is no need for enumeration, I wrote them down because the list is indeed short. For each scenario, there are 3C1=3, 3C2=3, possibilities, respectively.
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Re: Five balls of different colors are to be placed in three [#permalink]
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VeritasPrepKarishma wrote: avaneeshvyas wrote: Five balls of different colors are to be placed in three different boxes such that any box contains at least 1 ball . What is the maximum number of different ways in which this can be done?
A. 60 B. 90 C. 120 D. 150 E. 180
Please provide a small note of explanation for all the combinations used in the solution. I did it the way walker did since we can avoid enumerating by using that method i.e. we avoid (3, 1, 1), (2, 2, 1) etc. When required to enumerate, there is a chance that we might forget one of the possible combinations (though in this question it doesn't matter much since number of balls is small). Do visit that solution properly too. In this question, consider it an exercise in lots of PnC concepts. In how many ways can you distribute 5 different balls in 3 different boxes such that a box may get no ball? (we will later subtract those cases in which one or two boxes are empty) Each ball can be placed in any of the 3 boxes so there are 3 ways to place a ball. Since there are 5 balls, we can place all the balls in 3*3*3*3*3 = 3^5 ways Only one box empty: In how many ways can you distribute the 5 different balls in 2 different boxes such that both boxes have at least one ball? (we want to find the cases in which one and only one box will be empty) Select the empty box in 3 ways. Each ball can be placed in any of the 2 boxes. So 5 balls can be placed in 2*2*2*2*2 = 2^5 ways. Out of 2^5, subtract those cases where the balls are all in 1 box only. This happens in 2 ways since you can select an empty box again out of the two in 2 ways. Total number of ways = 3*(2^5  2). 2 boxes empty: In how many ways can you distribute the 5 balls in only 1 box? In 3 ways since you select a box in 3 ways. You can distribute 5 balls in 3 boxes such that no box is empty in 3^5  3*(2^5  2)  3 = 243  90  3 = 150 ways For a discussion on distribution of objects, check out these posts too: http://www.veritasprep.com/blog/2011/12 ... 93part1/http://www.veritasprep.com/blog/2011/12 ... spartii/Can someone please explain the text I have marked in Red?



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Re: Five balls of different colors are to be placed in three [#permalink]
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29 Aug 2013, 03:42
theGame001 wrote: VeritasPrepKarishma wrote: avaneeshvyas wrote: Five balls of different colors are to be placed in three different boxes such that any box contains at least 1 ball . What is the maximum number of different ways in which this can be done?
A. 60 B. 90 C. 120 D. 150 E. 180
Please provide a small note of explanation for all the combinations used in the solution. I did it the way walker did since we can avoid enumerating by using that method i.e. we avoid (3, 1, 1), (2, 2, 1) etc. When required to enumerate, there is a chance that we might forget one of the possible combinations (though in this question it doesn't matter much since number of balls is small). Do visit that solution properly too. In this question, consider it an exercise in lots of PnC concepts. In how many ways can you distribute 5 different balls in 3 different boxes such that a box may get no ball? (we will later subtract those cases in which one or two boxes are empty) Each ball can be placed in any of the 3 boxes so there are 3 ways to place a ball. Since there are 5 balls, we can place all the balls in 3*3*3*3*3 = 3^5 ways Only one box empty: In how many ways can you distribute the 5 different balls in 2 different boxes such that both boxes have at least one ball? (we want to find the cases in which one and only one box will be empty) Select the empty box in 3 ways. Each ball can be placed in any of the 2 boxes. So 5 balls can be placed in 2*2*2*2*2 = 2^5 ways. Out of 2^5, subtract those cases where the balls are all in 1 box only. This happens in 2 ways since you can select an empty box again out of the two in 2 ways. Total number of ways = 3*(2^5  2). 2 boxes empty: In how many ways can you distribute the 5 balls in only 1 box? In 3 ways since you select a box in 3 ways. You can distribute 5 balls in 3 boxes such that no box is empty in 3^5  3*(2^5  2)  3 = 243  90  3 = 150 ways For a discussion on distribution of objects, check out these posts too: http://www.veritasprep.com/blog/2011/12 ... 93part1/http://www.veritasprep.com/blog/2011/12 ... spartii/Can someone please explain the text I have marked in Red? You have 3 boxes. YOu want to keep exactly one empty. This means the other two must have at least one ball. Say the boxes are A, B and C. In how many ways can you choose to have exactly one box empty? A empty, B and C non empty B empty, A and C non empty C empty, A and B non empty Basically you can select the box you will keep empty in 3 ways. Now you have to put 5 balls in 2 boxes (say e.g. B and C) such that both boxes have at least one ball. For each ball, there are two options: box B or box C So you can put in 5 balls in 2*2*2*2*2 = 2^5 ways. But this includes cases where all ball are put in one box only. How many such cases are there? All balls could go in box B and box C could be empty. All balls could go in box C and box B would be empty. Only these two cases are possible. So out of the 2^5 cases, you need to remove these two cases because you must have at least one ball in each of the two boxes, B and C. No of ways in which you have all balls in exactly two boxes = 3*(2^5  2)
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How many ways can you distribute 5 marbles in 3 identical ba [#permalink]
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30 Aug 2013, 13:18
How many ways can you distribute 5 marbles in 3 identical baskets such that each basket gets at least 1 marble.
I am trying to solve this problems using as many approaches:
The correct answer is: we can pick distribute the marbles in two such ways: 311 or 221. For the 311. We choose 3 marbles for the 1st basket. (5C3). 1 for the 2nd basket (2C1) and from the remaining marble (1C1). We then divide by 2!. I know it has something to do with the the fact we are putting 1 marble into the 2 baskets but what is the intuition?? How do you if you have to divide by n!?? This is what I am not getting. After we have (5C3)(2C1)(1C1) does the problem become... " How many ways can we rearrange the digits 311?" Ans: 3!/(2!1!)
If I choose 3 people out of 5 for a soccer game its 5*4*3/3!. I divide by 3! Because order does not matter. How does this translate to the above problem?



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Re: How many ways can you distribute 5 marbles in 3 identical ba [#permalink]
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30 Aug 2013, 13:40
n1Cr1 = 51C31 = 4C2 = 6 There are direct formula: 1. Distribute n identical objects to r people, such that each can receive 0,1,2,3...n objects. (n+r1)C(r1) 2. Distribute n identical objects to r people, such that each receive at least one object. (n1)C(r1) This is what I used in above case, identical baskets sounds tricky, but we can place baskets in series 1 2 3.
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Re: How many ways can you distribute 5 marbles in 3 identical ba [#permalink]
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30 Aug 2013, 17:42
alphabeta1234 wrote: How many ways can you distribute 5 marbles in 3 identical baskets such that each basket gets at least 1 marble.
I am trying to solve this problems using as many approaches:
The correct answer is: we can pick distribute the marbles in two such ways: 311 or 221. For the 311. We choose 3 marbles for the 1st basket. (5C3). 1 for the 2nd basket (2C1) and from the remaining marble (1C1). We then divide by 2!. I know it has something to do with the the fact we are putting 1 marble into the 2 baskets but what is the intuition?? How do you if you have to divide by n!?? This is what I am not getting. After we have (5C3)(2C1)(1C1) does the problem become... " How many ways can we rearrange the digits 311?" Ans: 3!/(2!1!)
If I choose 3 people out of 5 for a soccer game its 5*4*3/3!. I divide by 3! Because order does not matter. How does this translate to the above problem? Hi, You see the constraint that each basket has to have at least 1 marble. So put 1 marble in each basket. We have 2 marbles remaining. This can be distibuted among 2 baskets or put in just 1 basket. The former can be done in 3C2 ways and the latter in 3C1 ways. So there are totally 6 possible ways you can do this. I assume the marbles are identical. So you do not have to consider 5C3.
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Re: Five balls of different colors are to be placed in three [#permalink]
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14 Nov 2013, 17:51
Hi i did it that way and i don't understand why i'm wrong: We can have two distributions: 1. 311 one box gets three balls, the 2 other boxes only 1 It is equivalent as to make 3 groups ordered out of 5 objects (as there are 3 different boxes) So 5C3 * 2C1 * 1C1 = 20 2. 122 one box gets one ball and the remaining two boxes get two balls each. It is equivalent as to make 3 groups ordered out of 5 objects (as there are 3 different boxes) So 5C1 * 4C2 * 2C2 = 30 so 50 solutions..



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Re: Five balls of different colors are to be placed in three [#permalink]
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14 Nov 2013, 21:18
oss198 wrote: Hi i did it that way and i don't understand why i'm wrong: We can have two distributions: 1. 311 one box gets three balls, the 2 other boxes only 1 It is equivalent as to make 3 groups ordered out of 5 objects (as there are 3 different boxes) So 5C3 * 2C1 * 1C1 = 20 2. 122 one box gets one ball and the remaining two boxes get two balls each. It is equivalent as to make 3 groups ordered out of 5 objects (as there are 3 different boxes) So 5C1 * 4C2 * 2C2 = 30 so 50 solutions.. The boxes are different B1, B2, B3 When you split the balls 311, which box gets 3 balls? You first select a box which gets 3 balls. You can do this in 3 ways  select B1 or B2 or B3. Now select 3 of the 5 balls for this box in 5C3 ways. Now you have 2 balls and 2 boxes one for each ball so you can distribute them in 2 ways. This gives 3*5C3*2 = 60. Same problem for 221 case. You multiply by another 3 and you get 3*30 = 90 Total 60+90 = 150
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