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In how many ways can we put 4 different balls in 3 different boxes whe [#permalink]
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In how many ways can we put 4 different balls in 3 different boxes when any box can contain any number of balls?A. 80B. 81C. 64D. 63E. 82
Originally posted by geetgmat on 28 Sep 2016, 09:40. Renamed the topic and edited the question.
Re: In how many ways can we put 4 different balls in 3 different boxes whe [#permalink]
geetgmat wrote: In how many ways can we put 4 different balls in 3 different boxes when any box can contain any number of balls?A. 80B. 81C. 64D. 63E. 82 Take the task of distributing the 4 different balls and break it into stages. Stage 1: Select a box for the 1st ball to go into. There are 3 available boxes, so we can complete stage 1 in 3 ways Stage 2: Select a box for the 2nd ball to go into. There are 3 available boxes, so we can complete stage 2 in 3 ways Stage 3: Select a box for the 3rd ball to go into. There are 3 available boxes, so we can complete stage 3 in 3 ways Stage 4: Select a box for the 4th ball to go into. There are 3 available boxes, so we can complete stage 4 in 3 ways By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus distribute all 4 balls) in (3)(3)(3)(3)(4) ways (= 81 ways) Answer: RELATED VIDEO_________________
Brent Hanneson – Creator of gmatprepnow.comI’ve spent the last 20 years helping students overcome their difficulties with GMAT math, and the biggest thing I’ve learned is… Many students fail to maximize their quant score NOT because they lack the skills to solve certain questions but because they don’t understand what the GMAT is truly testing - Learn more
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Re: In how many ways can we put 4 different balls in 3 different boxes whe [#permalink]
geetgmat wrote: In how many ways can we put 4 different balls in 3 different boxes when any box can contain any number of balls?8081646382Answer: B.Can someone please tell me why 4x4x4=64 is not the right answer?There are 4 ways to fill the 1st box, and 4 ways to the 2nd box and 4 ways to fill the 3rd box..since it says that any box can contain any number of balls..i assumed that repetition is allowed.Kindly explain. Thanks alot!!! NOTE : The number of permutations/arrangments of n things , taken r at ta time when each item may be repeated once, twice...up to r times in any arrangement is \(n^r\) ways. The first box can be filled in n ways, the second box can be filled in n ways(even though the first box is filled with one item, the same item can be used for filling the second box also because repetition is allowed), the third box can also be filled in n ways...The rth box can be filled in n ways..Now all the r boxes together can be filled in (n*n*n*....r times ) ie. \(n^r\) ways.So \(3^4\) ways..Hope it clears..
Re: In how many ways can we put 4 different balls in 3 different boxes whe [#permalink]
geetgmat wrote: In how many ways can we put 4 different balls in 3 different boxes when any box can contain any number of balls?8081646382Answer: B.Can someone please tell me why 4x4x4=64 is not the right answer? There are 4 ways to fill the 1st box, and 4 ways to the 2nd box and 4 ways to fill the 3rd box..since it says that any box can contain any number of balls..i assumed that repetition is allowed. Kindly explain.Thanks alot!!! "There are 4 ways to fill the 1st box" This assumes that the box needs to be filled. No requirement for any one box to contain ANY balls. On the other hand, every ball needs to be placed somewhere. So, we need to examine the number of ways to place each ball. See my solution above.Cheers,Brent _________________
Brent Hanneson – Creator of gmatprepnow.comI’ve spent the last 20 years helping students overcome their difficulties with GMAT math, and the biggest thing I’ve learned is… Many students fail to maximize their quant score NOT because they lack the skills to solve certain questions but because they don’t understand what the GMAT is truly testing - Learn more
Intern Joined: 03 Feb 2017 Posts: 21
Re: In how many ways can we put 4 different balls in 3 different boxes whe [#permalink] Since there are 3 boxes, we have 3 possibilities for the first ball 3 possibilities for the second ball 3 possibilities for the third ball 3 possibilities for the fourth ball So, total number of possibilities is 3*3*3*3 = 81
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Re: In how many ways can we put 4 different balls in 3 different boxes whe [#permalink]
GMATPrepNow wrote: geetgmat wrote: In how many ways can we put 4 different balls in 3 different boxes when any box can contain any number of balls?8081646382Answer: B.Can someone please tell me why 4x4x4=64 is not the right answer? There are 4 ways to fill the 1st box, and 4 ways to the 2nd box and 4 ways to fill the 3rd box..since it says that any box can contain any number of balls..i assumed that repetition is allowed. Kindly explain.Thanks alot!!! "There are 4 ways to fill the 1st box" This assumes that the box needs to be filled. No requirement for any one box to contain ANY balls. On the other hand, every ball needs to be placed somewhere. So, we need to examine the number of ways to place each ball. See my solution above.Cheers,Brent How could people not give Kudos for such valuable information! Members please appreciate valuable contribution such as this one by giving Kudos! Thanks GMATPrepNow. _________________
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Re: In how many ways can we put 4 different balls in 3 different boxes whe [#permalink] Another way to think about it by considering the number of balls in each box. Situation 1Box 1: 4 balls | 0 balls | 0 ballsBox 2: 0 balls | 4 balls | 0 balls Box 3: 0 balls | 0 balls | 4 balls So situation 1 has 3 different possibilities. Similarly, situation 2, 3, and 4 have 3 different possibilities each. Situation 2Box 1: 3 ballsBox 2: 1 balls Box 3: 0 balls Situation 3Box 1: 2 ballsBox 2: 2 balls Box 3: 0 balls Situation 4Box 1: 2 ballsBox 2: 1 balls Box 3: 1 balls Thus the total number of ways in which 4 different balls can be put in 3 different boxes when any box can contain any number of balls is3 x 3 x 3 x 3 = 81 _________________
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SVP Joined: 24 Nov 2016 Posts: 1806 Location: United States
In how many ways can we put 4 different balls in 3 different boxes whe [#permalink]
geetgmat wrote: In how many ways can we put 4 different balls in 3 different boxes when any box can contain any number of balls?A. 80B. 81C. 64D. 63 E. 82 given: 4 dif balls, 3 dif boxes, any number per box;\(4•4•4=81\)\({400}={3!/2!}•4c4=3\) ("/2!" because 00 are identical)\({310}={3!}•4c3=6•4=24\)\({220}={3!}•4c2/2!=6•3=18\) ("/2!" because 22 is double counting)\({211}={3!}•4c2=6•6=36\)\({total}=3+24+18+36=81\)Answer (B)
Re: In how many ways can we put 4 different balls in 3 different boxes whe [#permalink] Hi Bunuel chetan2u, I was trying to use the divider method that dabaobao has used in https://gmatclub.com/forum/in-how-many-ways-can-5-apples-identical-be-distributed-among-4-child-206279.html Like this: BB|B|B. Therefore, 6!/(2!*4!). However, I am not getting the answer. Can you please advice where I am going wrong?
Math Expert Joined: 02 Aug 2009 Posts: 10604
Re: In how many ways can we put 4 different balls in 3 different boxes whe [#permalink]
VIVA1060 wrote: Hi Bunuel chetan2u, I was trying to use the divider method that dabaobao has used in https://gmatclub.com/forum/in-how-many-ways-can-5-apples-identical-be-distributed-among-4-child-206279.html Like this: BB|B|B. Therefore, 6!/(2!*4!). However, I am not getting the answer. Can you please advice where I am going wrong? That formula is for identical things but here we have different things. Say we have 3 boxes A,B and C and ball w,x,y, and zThE formula gives A- 3 boxes, B- 1 box and C-0 box as one way. But the actuals where balls are different A- wxy, B-z, C-0 and A-xyz, B-w,C-0 will be treated differently. _________________
Re: In how many ways can we put 4 different balls in 3 different boxes whe [#permalink]
exc4libur wrote: geetgmat wrote: In how many ways can we put 4 different balls in 3 different boxes when any box can contain any number of balls?A. 80B. 81C. 64D. 63 E. 82 given: 4 dif balls, 3 dif boxes, any number per box;\(4•4•4=81\)\({400}={3!/2!}•4c4=3\) ("/2!" because 00 are identical)\({310}={3!}•4c3=6•4=24\)\({220}={3!}•4c2/2!=6•3=18\) ("/2!" because 22 is double counting) \({211}={3!}•4c2=6•6=36\)\({total}=3+24+18+36=81\)Answer (B) Hey! can you help me in understanding this case : \({220}={3!}•4c2/2!=6•3=18\) ("/2!" because 22 is double counting) why you divided by 2! in the case 220. The cases will be 220 , 202, 022 and all these are different.Thanks in advance _________________
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Re: In how many ways can we put 4 different balls in 3 different boxes whe [#permalink] Hello from the GMAT Club BumpBot!Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________
Re: In how many ways can we put 4 different balls in 3 different boxes whe [#permalink] 18 Feb 2022, 04:37 |
How many ways 10 different balls can be put into 2 different boxes so that each box has at least one ball?
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