In how many ways 5 identical balls can be distributed into 3 different boxes so that no box remains empty?
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In how many ways can five identical balls be distributed in three [#permalink]
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Question Stats: Hide Show timer StatisticsIn how many ways can five identical balls be distributed in three different boxes?A)15B)18C)20D)21 E)24
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Re: In how many ways can five identical balls be distributed in three [#permalink] Number of balls in 3 boxes are A, B and C, where A,B and C are non-negative integers.A+B+C=5A|B|C (Total partitions= 2)Number of ways to arrange 5 similar balls and 2 partitions= (5+2)C2= 7C2 = 21 kawal27 wrote: In how many ways can five identical balls be distributed in three different boxes?A)15B)18C)20D)21 E)24
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Re: In how many ways can five identical balls be distributed in three [#permalink]
There are 2 important equations for these types of arrangementsIf the number of non negative integral solutions for the equation \(x_1 + x_2 +.. + x_n = n\), then the number of ways the distribution can be done is: (i) \( ^{n + r - 1} C _ {r - 1}\). In this case, value of any variable can be zero.(ii) \(^{n - 1} C _ {r - 1}\). In this case, minimum value for any variable is 1.Given that: n = 5 balls, r = 3 boxes and some of the boxes can have zero balls (as nothing is specified that each box should have at least 1 ball).Therefore the total number of ways = \(^{5 + 3 - 1} C _ {3 - 1}= ^7C_2 = \frac{7 * 6}{2 * 1} = 21\) Option D Arun Kumar _________________
Manager Joined: 26 Jul 2020 Posts: 219
Re: In how many ways can five identical balls be distributed in three [#permalink] I did not know the formula, so I did it with a different approach. I don't know where I went wrong.Arrangement 1: 1 box has all 5 balls. So, arranging "500" in different ways 3!/2! (since 0 is repeated) = 3Arrangement 2: 1 box has 4 balls, 1 other box has 1. Third has 0. So, arranging "410" in different ways 3! = 6Arrangement 3: 1 box has 3 balls, other 2 boxes have 1 each. So, arranging "311" in different ways 3!/2! (since 1 is repeated) = 3Arrangement 4: 1 box has 2 balls, second box has 2 and third has 1. So, arranging "221" in different ways 3!/2! (since 2 is repeated) = 3 I got a total of 15. What combination did I miss?
Manager Joined: 26 Jul 2020 Posts: 219
Re: In how many ways can five identical balls be distributed in three [#permalink]
aviddd wrote: I did not know the formula, so I did it with a different approach. I don't know where I went wrong.Arrangement 1: 1 box has all 5 balls. So, arranging "500" in different ways 3!/2! (since 0 is repeated) = 3Arrangement 2: 1 box has 4 balls, 1 other box has 1. Third has 0. So, arranging "410" in different ways 3! = 6Arrangement 3: 1 box has 3 balls, other 2 boxes have 1 each. So, arranging "311" in different ways 3!/2! (since 1 is repeated) = 3Arrangement 4: 1 box has 2 balls, second box has 2 and third has 1. So, arranging "221" in different ways 3!/2! (since 2 is repeated) = 3 I got a total of 15. What combination did I miss? Darn it. I realized as soon as I hit Submit. I missed 320 combination. Which gives me the missing 6.
Manager Joined: 26 Jul 2020 Posts: 219
Re: In how many ways can five identical balls be distributed in three [#permalink]
CrackVerbalGMAT wrote: There are 2 important equations for these types of arrangementsIf the number of non negative integral solutions for the equation \(x_1 + x_2 +.. + x_n = n\), then the number of ways the distribution can be done is: (i) \( ^{n + r - 1} C _ {r - 1}\). In this case, value of any variable can be zero.(ii) \(^{n - 1} C _ {r - 1}\). In this case, minimum value for any variable is 1.Given that: n = 5 balls, r = 3 boxes and some of the boxes can have zero balls (as nothing is specified that each box should have at least 1 ball).Therefore the total number of ways = \(^{5 + 3 - 1} C _ {3 - 1}= ^7C_2 = \frac{7 * 6}{2 * 1} = 21\) Option D Arun Kumar Can you give more examples where this equation can be used?
In how many ways can five identical balls be distributed in three [#permalink]
kawal27 wrote: In how many ways can five identical balls be distributed in three different boxes?A)15B)18C)20D)21 E)24 Since all five answer choices are relatively small, we can probably list and count all of the possible outcomes in under 2 minutes. Let's list the outcomes the following way: ABC such that:The first number is the number of balls in box A The second number is the number of balls in box B The third number is the number of balls in box CWe get: 500410401320302311230203221212140104113131122050005041014032023DONE!There are 21 outcomesAnswer: D _________________
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Re: In how many ways can five identical balls be distributed in three [#permalink] Hi BrentGMATPrepNow,Thanks for a great explanation. Just a small typo in the list 132 instead of 131, please correct it.
Re: In how many ways can five identical balls be distributed in three [#permalink]
ravishekar wrote: Hi BrentGMATPrepNow,Thanks for a great explanation. Just a small typo in the list 132 instead of 131, please correct it. Good catch, thanks! I've edited my response. Kudos for you!! _________________
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 five identical balls be distributed in three [#permalink]
kawal27 wrote: In how many ways can five identical balls be distributed in three different boxes?A)15B)18C)20D)21 E)24 The following is called the SEPARATOR method.Five identical balls are to be separated into -- at most -- 3 groupings.Thus, we need five balls and two separators:OO|OO|OEvery arrangement of the elements above represents one way to distribute the 5 balls among three boxes A, B and C:OO|OO|O = A gets 2 balls, B gets 2 balls, C gets 1 ball.OO||OOO = A gets 2 balls, B gets 0 balls, C gets 3 balls.OOOOO|| = A gets all 5 balls.And so on.To count all of the possible distributions, we simply need to count the number of ways to arrange the 7 elements above (the 5 identical balls and the 2 identical separators).The number of ways to arrange 7 elements = 7!.But when an arrangement includes identical elements, we must divide by the number of ways each set of identical arrangements can be arranged.The reason:When the identical elements swap positions, the arrangement doesn't change, reducing the total number of unique arrangements.Here, we must divide by 5! to account for the 5 identical balls and by 2! to account for the 2 identical separators:\(\frac{7!}{5!2!} = 21\) _________________
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Re: In how many ways can five identical balls be distributed in three [#permalink] Asked: In how many ways can five identical balls be distributed in three different boxes?BBBBB||Number of ways = 7!/5!2! = 21IMO D _________________
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Re: In how many ways can five identical balls be distributed in three [#permalink]
kawal27 wrote: In how many ways can five identical balls be distributed in three different boxes?A)15B)18C)20D)21 E)24 total ways to distribute five identical balls be distributed in three different boxes5+3-1 C 3-1 7c2 ; 21 waysoption D
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Re: In how many ways can five identical balls be distributed in three [#permalink] There are 2 ways to solve this question. 1. If you know the formula of "Theory of Partitioning", n+r-1 C r-1, where n are the number of objects and r will be the number of partitions. Always remember, 'r' will always be one less than the number of divisions to be done. 2. For the ones who do not want to remember the formula : We know that there are 5 IDENTICAL objects which need to be partitioned in 3 boxes,i.e. 2 separations. Total objects= 7. Hence 7! / 5!2! Posted from my mobile device
Re: In how many ways can five identical balls be distributed in three [#permalink] Asked: In how many ways can five identical balls be distributed in three different boxes?BBBBB||There are 5 identical balls and 2 partitionsNumber of ways = 7!/5!2! = 21IMO D _________________
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Re: In how many ways can five identical balls be distributed in three [#permalink] 24 Nov 2021, 12:23 |
What is the number of ways of distributing 5 identical balls into 3 boxes so that no box remains empty?
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