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Four boys picked up 30 mangoes .In how many ways can they [#permalink]
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07 Aug 2010, 06:26
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Four boys picked up 30 mangoes .In how many ways can they divide them if all mangoes be identical?



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Re: solve these GMAT question [#permalink]
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07 Aug 2010, 07:39
Each Mango can be given to any one of the four people or in other words..1 mango can be divided into 4 ways...so all 30 can be divided in 4^30 ways..hops this helps



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Re: solve these GMAT question [#permalink]
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08 Aug 2010, 04:51
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likithae wrote: Four boys picked up 30 mangoes .In how many ways can they divide them if all mangoes be identical?
help me to solve ......... 1. If boys can get zero mangoes them the answer would be  \((30+41)C(41)=33C3\):Consider 30 Mangoes: ****************************** and 3 separators . Permutations of these 33 symbols out of which 30 *'s and 3 's are identical is \(\frac{33!}{3!30!}\), or written in another way \(C^3_{33}\). Each permutation will mean one particular distribution of 30 mangoes among 4 boys: ****************************** first boy gets all mangoes; ****************************** first, second and third boys get 1 mango each, and fourth gets 27; ***************************** first gets one mango, second boy gets zero, third boy get 1 and fourth gets 28; And so on. 2. If boys should get at least one mango then the answer would be  \((26+41)C(41)=29C3\) (basically we are distributing 26 mangoes):The same as above: we should jut give 1 mango to each boy and then distribute 26 mangoes left as in previous case. 26 Mangoes: ************************** and 3 separators . Permutations of these 29 symbols out of which 26 *'s and 3 's are identical is \(\frac{29!}{3!26!}\), or written in another way \(C^3_{29}\). Again each permutation will mean one particular distribution of 26 mangoes among 4 boys. Similar problems: voucher98225.html?hilit=separatorsintegerslessthan85291.html?hilit=identical#p710836Direct formula: The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0,1,2 or more items is \(n+r1C_{r1}\).
The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is \(n1C_{r1}\).P.S. 4^30 would be the answer if all mangoes were different but we are told that they are identical. Hope it helps.
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Re: solve these GMAT question [#permalink]
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08 Aug 2010, 09:24
Great analysis. I keep getting confused on such problems as to how to divide it up between so many persons. As the previous post suggested 4^30 seemed to make sense but then they are identical, so that is not correct. Let me ask this: 4^30 is correct number of distributing 30 different mangoes between 4 persons, if each gets 0,1,2... What if each has to get at least 1, 2, 3.. what would be the answer in that case with different mangoes?
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Re: solve these GMAT question [#permalink]
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02 Jul 2011, 18:31
Excellent explanation from Bunuel  "stars and bars" method unleashed!
Basically "indistinguishable" balls into "distinguishable" urns...
30 "indistinguishable" mangoes into 4 "distinguishable" boys..
33C30
ANS: 5456



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Re: solve these GMAT question [#permalink]
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25 May 2013, 23:38
mainhoon wrote: Great analysis. I keep getting confused on such problems as to how to divide it up between so many persons. As the previous post suggested 4^30 seemed to make sense but then they are identical, so that is not correct. Let me ask this:
4^30 is correct number of distributing 30 different mangoes between 4 persons, if each gets 0,1,2... What if each has to get at least 1, 2, 3.. what would be the answer in that case with different mangoes? Hey can you explain me the logic for identical and nonidentical stuff? I am good with explanation above but 4^30 is also making sense to me. What makes it different from 33 C 3
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Re: solve these GMAT question [#permalink]
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26 May 2013, 03:37
mainhoon wrote: Great analysis. I keep getting confused on such problems as to how to divide it up between so many persons. As the previous post suggested 4^30 seemed to make sense but then they are identical, so that is not correct. Let me ask this:
4^30 is correct number of distributing 30 different mangoes between 4 persons, if each gets 0,1,2... What if each has to get at least 1, 2, 3.. what would be the answer in that case with different mangoes? Hi Bunnel,, Can you please answer the above query?



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Re: solve these GMAT question [#permalink]
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03 Sep 2013, 06:16
Bunuel wrote: likithae wrote: Four boys picked up 30 mangoes .In how many ways can they divide them if all mangoes be identical?
help me to solve ......... 1. If boys can get zero mangoes them the answer would be  \((30+41)C(41)=33C3\):Consider 30 Mangoes: ****************************** and 3 separators . Permutations of these 33 symbols out of which 30 *'s and 3 's are identical is \(\frac{33!}{3!30!}\), or written in another way \(C^3_{33}\). Each permutation will mean one particular distribution of 30 mangoes among 4 boys: ****************************** first boy gets all mangoes; ****************************** first, second and third boys get 1 mango each, and fourth gets 27; ***************************** first gets one mango, second boy gets zero, third boy get 1 and fourth gets 28; And so on. 2. If boys should get at least one mango then the answer would be  \((26+41)C(41)=29C3\) (basically we are distributing 26 mangoes):The same as above: we should jut give 1 mango to each boy and then distribute 26 mangoes left as in previous case. 26 Mangoes: ************************** and 3 separators . Permutations of these 29 symbols out of which 26 *'s and 3 's are identical is \(\frac{29!}{3!26!}\), or written in another way \(C^3_{29}\). Again each permutation will mean one particular distribution of 26 mangoes among 4 boys. Similar problems: voucher98225.html?hilit=separatorsintegerslessthan85291.html?hilit=identical#p710836Direct formula: The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0,1,2 or more items is \(n+r1C_{r1}\).
The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is \(n1C_{r1}\).P.S. 4^30 would be the answer if all mangoes were different but we are told that they are identical. Hope it helps. Bunuel, Why is the same formula \(n+r1C_{r1}\) used for number of ways of dividing n identical items among r persons whom can receive can receive 0,1,2 or more items . Are you saying the same formula is used even if each person can receive 2 or 3 or anyitem? Why isn't that being factored in to the formula? How many ways are there to distribute 10 mangoes among 4 kids if each kid must receive at least 1 mango? How many ways are there to distribute 10 mangoes among 4 kids if each kid must receive at least 2 mango? How many ways are there to distribute 10 mangoes among 4 kids if each kid must receive at least 3 mango? Are you saying this formula gives us the same answer for all these questions? (10+41)C(41)?



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Re: Four boys picked up 30 mangoes .In how many ways can they [#permalink]
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Re: Four boys picked up 30 mangoes .In how many ways can they [#permalink]
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Four boys picked up 30 mangoes .In how many ways can they [#permalink]
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28 Aug 2016, 06:45
Quote: Bunuel,
Why is the same formula \(n+r1C_{r1}\) used for number of ways of dividing n identical items among r persons whom can receive can receive 0,1,2 or more items . Are you saying the same formula is used even if each person can receive 2 or 3 or anyitem? Why isn't that being factored in to the formula?
How many ways are there to distribute 10 mangoes among 4 kids if each kid must receive at least 1 mango?
How many ways are there to distribute 10 mangoes among 4 kids if each kid must receive at least 2 mango?
How many ways are there to distribute 10 mangoes among 4 kids if each kid must receive at least 3 mango?
Are you saying this formula gives us the same answer for all these questions? (10+41)C(41)? I guess you already don't need a reply, but nevertheless: 1) 10 mangoes to 4 kids: We have 10 mangoes and 3 separators, so together 14 elements, then the number combinations is : \(\frac{13!}{3!10!}\) 2) 10 mangoes to 4 kids, if each one has to get at least 1 mango: First we give 1 mango to each, this leaves 6 mangoes, so we need to distribute this 6 mangoes to them => 6 mangoes and 3 separators: \(\frac{9!}{6!3!}\) 3) 10 mangoes to 4 kids, if one has to get at least 2 mangoes: First we give 2 mango to each, this leaves 2 mangoes, so we need to distribute this 2 mangoes to them => 2 mangoes and 3 separators: \(\frac{5!}{2!3!}\) 4) In order for each one to have 3 mangoes, the total number of mangoes has to be 12, so this option is not feasible




Four boys picked up 30 mangoes .In how many ways can they
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