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In how many ways can 5 different rings be worn in four [#permalink]
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02 Feb 2012, 16:16
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In how many ways can 5 different rings be worn in four particular fingers? (Some fingers may get more than one ring and some may get no rings.)
Can somebody explain?



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Re: 5 rings in four fingers [#permalink]
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02 Feb 2012, 16:23



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Re: 5 rings in four fingers [#permalink]
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02 Feb 2012, 16:44
I tried to understand the ringfinger arrangements issue form the link you posted, but the trail appears contradictory. Since that was in 2009, can you pls reconfirm were your Answers & Explanations correct in that trail.
5ringson4fingers86111.html?hilit=rings%20finger#p649557



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Re: 5 rings in four fingers [#permalink]
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02 Feb 2012, 16:51



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Re: In how many ways can 5 different rings be worn in four [#permalink]
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02 Feb 2012, 17:28
Many Thnx.
2 similar Qs.
1. In how many ways can 5 Idnetical fruits be distributed in 4 identical baskets?
Fruits are identical but baskets are also identical so we cannot apply 8C3 from n+r1 C r1? Can we say that to remove the duplications of baskets we should divide 8C3 by 4! ?
2. In how many ways can 5 different fruits be distributed in 4 identical baskets? Is this like making 4 groups from 5 different fruits so = 5!



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Re: In how many ways can 5 different rings be worn in four [#permalink]
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02 Feb 2012, 18:24
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docabuzar wrote: Many Thnx.
2 similar Qs.
1. In how many ways can 5 Idnetical fruits be distributed in 4 identical baskets?
Fruits are identical but baskets are also identical so we cannot apply 8C3 from n+r1 C r1? Can we say that to remove the duplications of baskets we should divide 8C3 by 4! ?
2. In how many ways can 5 different fruits be distributed in 4 identical baskets? Is this like making 4 groups from 5 different fruits so = 5! I think you are overthinking the concept by introducing the "identical basket" situation. For the first case if, for example, there is no difference between the following scenarios: {5000} and {0500}, then you can manually write down all possible cases: {5000}, {4100}, {3110}, {3200}, {2210}, {2111}. So the answer for the first question as you stated would be 6. Don't overcomplicate it: the GMAT combination/probability questions are fairly straightforward and no need to waste time on the problems you will never see on the test. If the questions were: 1. In how many ways can 5 identical fruits be distributed in 4 different baskets?Consider five stars and three bars: *****, where stars represent fruits and three bars (one less than the baskets) will help us to divide them among the baskets. How many permutations (arrangements) of these 5+3=8 symbols are there? Permutation of 8 symbols out of which 5 * and 3  are identical is 8!/(5!3!). Now, each arrangement will mean different scenario for fruit distribution. For example: ***** will mean that the first basket gets 2 fruits, the second, third and fourth 1 fruit. Or: **** will mean that the first basket gets 3 fruits, the second 1, and third and fourth none, and so on. Answer: 8!/(5!3!). 2. In how many ways can 5 different fruits be distributed in different 4 baskets?This one is easier each of the 5 different fruits has 4 choices, so total # of distribution is 4*4*4*4*4=4^5. Answer: 4^5. Check similar questions: howmanypositiveintegerslessthan10000aretherein85291.htmlcombinationstough108739.htmlsolvethesegmatquestion98701.htmlvoucher98225.htmlTheory on permutation and combinations: permutation86687.htmlDirect formula if needed: 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 \(n1C_{r1}\).Hope it helps.
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Re: In how many ways can 5 different rings be worn in four [#permalink]
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02 Feb 2012, 18:46
Many Thanks for the clarification. Both the original question in the post above with "Identical basket" & different basket were given in a post by veritasprep i think by Kbansal, I read & saved some time back.



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Re: In how many ways can 5 different rings be worn in four [#permalink]
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02 Feb 2012, 19:18
docabuzar wrote: Many Thanks for the clarification. Both the original question in the post above with "Identical basket" & different basket were given in a post by veritasprep i think by Kbansal, I read & saved some time back. OK. Below is the solution for your second question as it was originally stated. Notice that it's not that hard but it involves some lengthy calculations and that's the reason it's doubtful one encounter such problem on the test. In how many ways can 5 different fruits be distributed in 4 identical baskets?You can have 6 cases and each will have its own # of combinations: {5000} > 1 way (as there is no difference in which identical basket all the fruits will go); {4100} > \(C^4_5=5\): choosing which 4 fruits out of 5 will be together in one basket; {3110} > \(C^3_5=10\): choosing which 3 fruits out of 5 will be together in one basket (it doesn't matter in which 2 baskets the remaining 2 fruit sill go); {3200} > \(C^3_5=10\): choosing which 3 fruits out of 5 will be together in one basket (it doesn't matter in which 1 baskets the remaining 2 fruit sill go); {2210} > \(C^1_5*\frac{C^2_4}{2!}=15\): choosing which single fruit goes to one basket and \(\frac{C^2_4}{2!}\) is the # of ways to split 4 different fruits in two baskets when the order of the baskets doesn't matter; {2111} > \(C^2_5=10\): choosing which 2 fruits out of 5 will be together in one basket; Total: 1+5+10+10+15+10=51.
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Re: In how many ways can 5 different rings be worn in four [#permalink]
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03 Aug 2013, 04:47
isn't the answer:
choose 4 rings out of 6 which means C(6,4) but I can choose to wear any ring in any finger so 4!
so answer should be C(6,4)* 4!
Please correct me if I am wrong.



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Re: In how many ways can 5 different rings be worn in four [#permalink]
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29 Aug 2013, 22:35
Bunuel wrote: docabuzar wrote: Many Thnx.
2 similar Qs.
1. In how many ways can 5 Idnetical fruits be distributed in 4 identical baskets?
Fruits are identical but baskets are also identical so we cannot apply 8C3 from n+r1 C r1? Can we say that to remove the duplications of baskets we should divide 8C3 by 4! ?
2. In how many ways can 5 different fruits be distributed in 4 identical baskets? Is this like making 4 groups from 5 different fruits so = 5! I think you are overthinking the concept by introducing the "identical basket" situation. For the first case if, for example, there is no difference between the following scenarios: {5000} and {0500}, then you can manually write down all possible cases: {5000}, {4100}, {3110}, {3200}, {2210}, {2111}. So the answer for the first question as you stated would be 6. Don't overcomplicate it: the GMAT combination/probability questions are fairly straightforward and no need to waste time on the problems you will never see on the test. If the questions were: 1. In how many ways can 5 identical fruits be distributed in 4 different baskets?
Consider five stars and three bars: *****, where stars represent fruits and three bars (one less than the baskets) will help us to divide them among the baskets. How many permutations (arrangements) of these 5+3=8 symbols are there? Permutation of 8 symbols out of which 5 * and 3  are identical is 8!/(5!3!). Now, each arrangement will mean different scenario for fruit distribution. For example: ***** will mean that the first basket gets 2 fruits, the second, third and fourth 1 fruit. Or: **** will mean that the first basket gets 3 fruits, the second 1, and third and fourth none, and so on.Answer: 8!/(5!3!). 2. In how many ways can 5 different fruits be distributed in different 4 baskets?This one is easier each of the 5 different fruits has 4 choices, so total # of distribution is 4*4*4*4*4=4^5. Answer: 4^5. Check similar questions: howmanypositiveintegerslessthan10000aretherein85291.htmlcombinationstough108739.htmlsolvethesegmatquestion98701.htmlvoucher98225.htmlTheory on permutation and combinations: permutation86687.htmlDirect formula if needed: 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 \(n1C_{r1}\).Hope it helps. Hi, Why cant we use the same logic as used in the ring problem? How does using identical and different change the solution? I mean these are not word problems where the combination has to be unique? Thanks



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Re: In how many ways can 5 different rings be worn in four [#permalink]
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