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Zeynep (the lady of the rings) has 10 identical gold, 10 identical sil
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Updated on: 18 May 2016, 05:14
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Zeynep (the lady of the rings) has 10 identical gold, 10 identical silver and 10 identical bronze rings to wear in a particular ceremony. She will wear exactly one gold, one silver and one bronze (total of 3) rings on each of her 10 fingers such that for each finger she will wear one gold ring at the bottom, closest to the knuckle; then one silver ring; then one bronze ring in the most distal position. (According to the rituals of the ceremony this exact arrangement is required on each of her ten fingers.) In how many ways can she execute the processing of putting these 30 rings on her 10 (distinguishable) fingers, so that she ends up with this fixed outcome? (Example of a possible order: a gold ring to the left index finger, then another gold ring to the right little finger, then another gold ring to the right middle finger, then a silver ring to left index finger, then a bronze ring to left index finger... and so on.) A) \(\frac{(10!)^3}{2^{10}}\) B) \(\frac{(10!)^3}{3^{10}}\) C) \(\frac{30!}{2^{10}}\) D) \(\frac{30!}{3^{10}}\) E) \(\frac{30!}{6^{10}}\) Source: ITU GREGMAT prep seminar. This question was presented as a possible very hard (top level) GREGMAT question. It's tricky but can be done under 2 minutes.
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Originally posted by leve on 16 May 2016, 15:07.
Last edited by leve on 18 May 2016, 05:14, edited 13 times in total.



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Re: Zeynep (the lady of the rings) has 10 identical gold, 10 identical sil
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Updated on: 17 May 2016, 16:47
leve wrote: Zeynep (the lady of the rings) has 10 identical gold, 10 identical silver and 10 identical bronze rings. (total of 30) She always wears exactly 3 rings on each of her 10 fingers in a particular sequence: first gold, then silver and last bronze. In how many different orders can she wear her 30 rings?
a) (10!)^3 / 2^10 b) (10!)^3 / 3^10 c) 30! / 2^10 d) 30! / 3^10 e) 30! / 6^10 Dear leve, I'm happy to respond. I found no evidence on the web for this question other than this posting here: is this question your own creation? If so, you should have tagged it as such. If not, exactly what is the source? Here is the issue. It seems, from my reading of the prompt, that there is no choice. In other words, on every finger, the estimable Ms. Zeynep will wear one gold ring at the bottom, closest to the knuckle; then one silver ring; then the bronze ring in the most distal position. It seems that exact arrangement is required on each of her ten fingers, which would mean the exact position of all 30 rings would be specified in the prompt. If you or the question author have something else in mind, I believe you or he need to make this quite explicit. The official questions are, among other things, masterpieces of clarity and precision in their wording. This is a lofty standard for which to strive. Please let me know if you have any questions. Mike
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Originally posted by mikemcgarry on 16 May 2016, 16:28.
Last edited by mikemcgarry on 17 May 2016, 16:47, edited 1 time in total.



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Re: Zeynep (the lady of the rings) has 10 identical gold, 10 identical sil
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17 May 2016, 03:08
Zeynep (the lady of the rings) has 10 identical gold, 10 identical silver and 10 identical bronze rings to wear in a particular ceremony. She will wear exactly one gold, one silver and one bronze (total of 3) rings on each of her 10 fingers such that for each finger she will wear one gold ring at the bottom, closest to the knuckle; then one silver ring; then one bronze ring in the most distal position. (According to the rituals of the ceremony this exact arrangement is required on each of her ten fingers.) In how many different orders can she wear her all 30 rings on her 10 (distinguishable) fingers?
(Example of a possible order: gold ring to the left index finger, then another gold ring to the right little finger, then another gold ring to the right middle finger, then a silver ring to left index finger, then a bronze ring to left index finger... and so on.)
a) (10!)^3 / 2^10 b) (10!)^3 / 3^10 c) 30! / 2^10 d) 30! / 3^10 e) 30! / 6^10
Source: ITU GREGMAT prep seminar.
This question was presented as a possible very hard (top level) GREGMAT question. It's tricky but can be done under 2 minutes.



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Re: Zeynep (the lady of the rings) has 10 identical gold, 10 identical sil
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17 May 2016, 11:02
leve wrote: Zeynep (the lady of the rings) has 10 identical gold, 10 identical silver and 10 identical bronze rings to wear in a particular ceremony. She will wear exactly one gold, one silver and one bronze (total of 3) rings on each of her 10 fingers such that for each finger she will wear one gold ring at the bottom, closest to the knuckle; then one silver ring; then one bronze ring in the most distal position. (According to the rituals of the ceremony this exact arrangement is required on each of her ten fingers.) In how many different orders can she wear her all 30 rings on her 10 (distinguishable) fingers?
(Example of a possible order: gold ring to the left index finger, then another gold ring to the right little finger, then another gold ring to the right middle finger, then a silver ring to left index finger, then a bronze ring to left index finger... and so on.)
a) (10!)^3 / 2^10 b) (10!)^3 / 3^10 c) 30! / 2^10 d) 30! / 3^10 e) 30! / 6^10
Source: ITU GREGMAT prep seminar.
This question was presented as a possible very hard (top level) GREGMAT question. It's tricky but can be done under 2 minutes. Dear leve, I'm happy to respond. My friend, I think I might be beginning to understand the confusion. If I understand correctly, what you are saying is that the final arrangement of the rings once all 30 are on her fingers is completely fixed. For the final arrangement, there is only one possibility, but that's not what you are asking. If I understand correctly, you intend to ask not In how many different orders can she wear her all 30 rings on her 10 (distinguishable) fingers?but rather In how many ways can she execute the processing of putting these 30 rings on her fingers, so that she ends up with this fixed outcome? Is this what the question is supposed to be asking? If it is, do you understand that this meaning was entirely opaque from your previous formulations? You don't want the number of final arrangements, but instead the number of different ways to get to that single final arrangement. The verb " wear" implied the finished state of being dressed: thus, I have only one way to wear two identical socks on my two feet. By contrast, the verb " putting on" implies the process of getting dressed: thus, I have two ways to put on two socks, because I could start with either foot. My friend, I gather that you encountered this problem in the past but don't have the exact wording, so you are trying to state the problem in your own wording. I think you are underestimating the high level of precision needed to formulate a Quantitative question in a rigorously unambiguous fashion. Never underestimate the tremendous value of the exact wording of a problem from a reputable source, because the slightest casual paraphrase can unintentionally compromise the logic of the entire question. I will forebear offering any solution until I am perfectly clear about what is being asked. Does all this make sense? Mike
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Re: Zeynep (the lady of the rings) has 10 identical gold, 10 identical sil
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17 May 2016, 11:42
mikemcgarry wrote: leve wrote: Zeynep (the lady of the rings) has 10 identical gold, 10 identical silver and 10 identical bronze rings to wear in a particular ceremony. She will wear exactly one gold, one silver and one bronze (total of 3) rings on each of her 10 fingers such that for each finger she will wear one gold ring at the bottom, closest to the knuckle; then one silver ring; then one bronze ring in the most distal position. (According to the rituals of the ceremony this exact arrangement is required on each of her ten fingers.) In how many different orders can she wear her all 30 rings on her 10 (distinguishable) fingers?
(Example of a possible order: gold ring to the left index finger, then another gold ring to the right little finger, then another gold ring to the right middle finger, then a silver ring to left index finger, then a bronze ring to left index finger... and so on.)
a) (10!)^3 / 2^10 b) (10!)^3 / 3^10 c) 30! / 2^10 d) 30! / 3^10 e) 30! / 6^10
Source: ITU GREGMAT prep seminar.
This question was presented as a possible very hard (top level) GREGMAT question. It's tricky but can be done under 2 minutes. Dear leve, I'm happy to respond. My friend, I think I might be beginning to understand the confusion. If I understand correctly, what you are saying is that the final arrangement of the rings once all 30 are on her fingers is completely fixed. For the final arrangement, there is only one possibility, but that's not what you are asking. If I understand correctly, you intend to ask not In how many different orders can she wear her all 30 rings on her 10 (distinguishable) fingers?but rather In how many ways can she execute the processing of putting these 30 rings on her fingers, so that she ends up with this fixed outcome? Is this what the question is supposed to be asking? If it is, do you understand that this meaning was entirely opaque from your previous formulations? You don't want the number of final arrangements, but instead the number of different ways to get to that single final arrangement. The verb " wear" implied the finished state of being dressed: thus, I have only one way to wear two identical socks on my two feet. By contrast, the verb " putting on" implies the process of getting dressed: thus, I have two ways to put on two socks, because I could start with either foot. My friend, I gather that you encountered this problem in the past but don't have the exact wording, so you are trying to state the problem in your own wording. I think you are underestimating the high level of precision needed to formulate a Quantitative question in a rigorously unambiguous fashion. Never underestimate the tremendous value of the exact wording of a problem from a reputable source, because the slightest casual paraphrase can unintentionally compromise the logic of the entire question. I will forebear offering any solution until I am perfectly clear about what is being asked. Does all this make sense? Mike Hi again Mike, Yes, the question is asking 'In how many ways can she execute the processing of putting these 30 rings on her 10 (distinguishable) fingers, so that she ends up with this fixed outcome? ' I revised the question as you suggested. I believe this time it is okay. Would you please check it?



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Re: Zeynep (the lady of the rings) has 10 identical gold, 10 identical sil
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17 May 2016, 16:56
leve wrote: Hi again Mike,
Yes, the question is asking 'In how many ways can she execute the processing of putting these 30 rings on her 10 (distinguishable) fingers, so that she ends up with this fixed outcome? '
I revised the question as you suggested. I believe this time it is okay. Would you please check it? Dear leveMy friend, the wording of the question is now clear. I don't know whether this is exactly as the GMAT would frame it, but at least it is clear what is being asked at this point. The only other suggestion I have would be to make the answer choices a little more fancy looking: A) \(\frac{(10!)^3}{2^{10}}\) B) \(\frac{(10!)^3}{3^{10}}\) C) \(\frac{30!}{2^{10}}\) D) \(\frac{30!}{3^{10}}\) E) \(\frac{30!}{6^{10}}\) By convention, the answer choice letters are typically listed as capital letters, not lower case. Also, professional math notation is always preferable to plaintext shortcuts. Finally, I will say, as intriguing a mathematical question as this is, it is strangely unGMATlike. You see, the business world is all about results. Once a result has been achieved, the route one took to achieving it rarely merits much attention. This counting problem is all about the different evanescent choices toward a fixed outcome: do you see how, in this way, it flies in the face of the priorities of the business world, however interesting it may be? Mike
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Re: Zeynep (the lady of the rings) has 10 identical gold, 10 identical sil
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18 May 2016, 05:30
mikemcgarry wrote: leve wrote: Hi again Mike,
Yes, the question is asking 'In how many ways can she execute the processing of putting these 30 rings on her 10 (distinguishable) fingers, so that she ends up with this fixed outcome? '
I revised the question as you suggested. I believe this time it is okay. Would you please check it? Dear leveMy friend, the wording of the question is now clear. I don't know whether this is exactly as the GMAT would frame it, but at least it is clear what is being asked at this point. The only other suggestion I have would be to make the answer choices a little more fancy looking: A) \(\frac{(10!)^3}{2^{10}}\) B) \(\frac{(10!)^3}{3^{10}}\) C) \(\frac{30!}{2^{10}}\) D) \(\frac{30!}{3^{10}}\) E) \(\frac{30!}{6^{10}}\) By convention, the answer choice letters are typically listed as capital letters, not lower case. Also, professional math notation is always preferable to plaintext shortcuts. Finally, I will say, as intriguing a mathematical question as this is, it is strangely unGMATlike. You see, the business world is all about results. Once a result has been achieved, the route one took to achieving it rarely merits much attention. This counting problem is all about the different evanescent choices toward a fixed outcome: do you see how, in this way, it flies in the face of the priorities of the business world, however interesting it may be? Mike Hi Mike, Thanks for your help. You are right about the business world. But we can consider this problem as a high level challenge problem. Since we are happy with the wording of the problem If it is possible can you erase the 'poor quality' tag from the tag list and unlock the question? I think this is a good challenging problem for especially advanced combinatorics practice.



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Re: Zeynep (the lady of the rings) has 10 identical gold, 10 identical sil
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18 May 2016, 05:34
leve wrote: mikemcgarry wrote: leve wrote: Hi again Mike,
Yes, the question is asking 'In how many ways can she execute the processing of putting these 30 rings on her 10 (distinguishable) fingers, so that she ends up with this fixed outcome? '
I revised the question as you suggested. I believe this time it is okay. Would you please check it? Dear leveMy friend, the wording of the question is now clear. I don't know whether this is exactly as the GMAT would frame it, but at least it is clear what is being asked at this point. The only other suggestion I have would be to make the answer choices a little more fancy looking: A) \(\frac{(10!)^3}{2^{10}}\) B) \(\frac{(10!)^3}{3^{10}}\) C) \(\frac{30!}{2^{10}}\) D) \(\frac{30!}{3^{10}}\) E) \(\frac{30!}{6^{10}}\) By convention, the answer choice letters are typically listed as capital letters, not lower case. Also, professional math notation is always preferable to plaintext shortcuts. Finally, I will say, as intriguing a mathematical question as this is, it is strangely unGMATlike. You see, the business world is all about results. Once a result has been achieved, the route one took to achieving it rarely merits much attention. This counting problem is all about the different evanescent choices toward a fixed outcome: do you see how, in this way, it flies in the face of the priorities of the business world, however interesting it may be? Mike Hi Mike, Thanks for your help. You are right about the business world. But we can consider this problem as a high level challenge problem. Since we are happy with the wording of the problem If it is possible can you erase the 'poor quality' tag from the tag list and unlock the question? I think this is a good challenging problem for especially advanced combinatorics practice. The topic is unlocked. Because the wording is not GMATlike the tags remain.
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Zeynep (the lady of the rings) has 10 identical gold, 10 identical sil
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23 Jun 2017, 00:33
leve wrote: Zeynep (the lady of the rings) has 10 identical gold, 10 identical silver and 10 identical bronze rings to wear in a particular ceremony. She will wear exactly one gold, one silver and one bronze (total of 3) rings on each of her 10 fingers such that for each finger she will wear one gold ring at the bottom, closest to the knuckle; then one silver ring; then one bronze ring in the most distal position. (According to the rituals of the ceremony this exact arrangement is required on each of her ten fingers.) In how many ways can she execute the processing of putting these 30 rings on her 10 (distinguishable) fingers, so that she ends up with this fixed outcome?
(Example of a possible order: a gold ring to the left index finger, then another gold ring to the right little finger, then another gold ring to the right middle finger, then a silver ring to left index finger, then a bronze ring to left index finger... and so on.)
A) \(\frac{(10!)^3}{2^{10}}\)
B) \(\frac{(10!)^3}{3^{10}}\)
C) \(\frac{30!}{2^{10}}\)
D) \(\frac{30!}{3^{10}}\)
E) \(\frac{30!}{6^{10}}\)
Source: ITU GREGMAT prep seminar.
This question was presented as a possible very hard (top level) GREGMAT question. It's tricky but can be done under 2 minutes. Okay..this was a really tough one for me..it took me half an hour to figure this one out. Here it goes.. Let's first list out what is given in the question.. We have 30 rings in total, and 10 fingers on which we have to wear them. There are 10 Gold, 10 Silver, and 10 Bronze rings. Every finger will have 3 rings worn in a fixed order, i.e. Gold, then Silver above it, and then Bronze on the top. We have been asked about the number of ways in which we can wear the rings like this. The question has given hints about the how the 'interactions' or 'movements' will happen. One of the example order of interactions that has been given to us is : Example of a possible order: a gold ring to the left index finger, then another gold ring to the right little finger, then another gold ring to the right middle finger, then a silver ring to left index finger, then a bronze ring to left index finger... and so on.To wear the 30 rings, we will have in total 30 movements of wearing some or the other ring on some or the other finger at a given point of time. The constraint that we have is  For each finger, Gold has to be worn first, then Silver, and finally Bronze. So we may fully cover any finger in however many movements, but this is how we have to wear them. I will give 2 more ways in which we can decide the order to wear. 1. Goldleft thumb, Goldright index finger, Silverright index finger, Goldleft ring finger, Bronzeright index finger,Goldright pinkie...and so on2. Goldright middle finger, Silverright middle finger, Goldleft index finger, Goldright pinkie, Silverleft index finger, Bronze right middle finger...and so on All ways will have a total of 30 interactions. Now we're gonna name every interaction and see. I'm going to call wearing a Gold ring on the 1st finger as \(G_1\), wearing a Silver ring on the 1st finger as \(S_1\), and wearing a Bronze ring on the 1st finger as \(B_1\), and so on for all the 10 unique fingers. I'm going to now write our 30 interactions in this subscript manner.. \(G_1\),\(G_7\),\(S_1\),\(G_5\),\(S_7\),\(B_1\),.....and so on as can be seen, each of these interactions is unique. So, in how many ways can we arrange 30 unique interactions? \(30!\) But we have a problem here, for every finger(let's say for 3rd finger).. \(G_3\) has to come before \(S_3\), and \(S_3\) has to come before \(B_3\) So this has to be our order \(G_3\)\(S_3\)\(B_3\) And this has to be true for all the fingers. What I mean to say is, for any of the \(30!\) arrangements, and for every finger, \(G_n\)\(S_n\)\(B_n\) has to be the order of wearing rings. These interactions may be separated by any number of interactions of other fingers, but this has to be the order. We know that in any of the \(30!\) arrangements, there will be interactions for each finger arranged in some order separated by any number of interactions.. for example.. \(G_n\)\(S_n\)\(B_n\) \(S_n\)\(G_n\)\(B_n\) \(B_n\)\(S_n\)\(G_n\)... \(3!\) = 6 ways. But we have to count only one of these orders \(G_n\)\(S_n\)\(B_n\) This has to be true for all the 10 fingers. So we have to divide \(30!\) by \(6\) for each finger to remove over counting. Final Answer \(\frac{30!}{6^{10}}\) E. mikemcgarry Bunuelcan something of this sort be expected on the GMAT?
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Re: Zeynep (the lady of the rings) has 10 identical gold, 10 identical sil
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23 Jun 2017, 01:56
ShashankDave wrote: leve wrote: Zeynep (the lady of the rings) has 10 identical gold, 10 identical silver and 10 identical bronze rings to wear in a particular ceremony. She will wear exactly one gold, one silver and one bronze (total of 3) rings on each of her 10 fingers such that for each finger she will wear one gold ring at the bottom, closest to the knuckle; then one silver ring; then one bronze ring in the most distal position. (According to the rituals of the ceremony this exact arrangement is required on each of her ten fingers.) In how many ways can she execute the processing of putting these 30 rings on her 10 (distinguishable) fingers, so that she ends up with this fixed outcome?
(Example of a possible order: a gold ring to the left index finger, then another gold ring to the right little finger, then another gold ring to the right middle finger, then a silver ring to left index finger, then a bronze ring to left index finger... and so on.)
A) \(\frac{(10!)^3}{2^{10}}\)
B) \(\frac{(10!)^3}{3^{10}}\)
C) \(\frac{30!}{2^{10}}\)
D) \(\frac{30!}{3^{10}}\)
E) \(\frac{30!}{6^{10}}\)
Source: ITU GREGMAT prep seminar.
This question was presented as a possible very hard (top level) GREGMAT question. It's tricky but can be done under 2 minutes. Okay..this was a really tough one for me..it took me half an hour to figure this one out. Here it goes.. Let's first list out what is given in the question.. We have 30 rings in total, and 10 fingers on which we have to wear them. There are 10 Gold, 10 Silver, and 10 Bronze rings. Every finger will have 3 rings worn in a fixed order, i.e. Gold, then Silver above it, and then Bronze on the top. We have been asked about the number of ways in which we can wear the rings like this. The question has given hints about the how the 'interactions' or 'movements' will happen. One of the example order of interactions that has been given to us is : Example of a possible order: a gold ring to the left index finger, then another gold ring to the right little finger, then another gold ring to the right middle finger, then a silver ring to left index finger, then a bronze ring to left index finger... and so on.To wear the 30 rings, we will have in total 30 movements of wearing some or the other ring on some or the other finger at a given point of time. The constraint that we have is  For each finger, Gold has to be worn first, then Silver, and finally Bronze. So we may fully cover any finger in however many movements, but this is how we have to wear them. I will give 2 more ways in which we can decide the order to wear. 1. Goldleft thumb, Goldright index finger, Silverright index finger, Goldleft ring finger, Bronzeright index finger,Goldright pinkie...and so on2. Goldright middle finger, Silverright middle finger, Goldleft index finger, Goldright pinkie, Silverleft index finger, Bronze right middle finger...and so on All ways will have a total of 30 interactions. Now we're gonna name every interaction and see. I'm going to call wearing a Gold ring on the 1st finger as \(G_1\), wearing a Silver ring on the 1st finger as \(S_1\), and wearing a Bronze ring on the 1st finger as \(B_1\), and so on for all the 10 unique fingers. I'm going to now write our 30 interactions in this subscript manner.. \(G_1\),\(G_7\),\(S_1\),\(G_5\),\(S_7\),\(B_1\),.....and so on as can be seen, each of these interactions is unique. So, in how many ways can we arrange 30 unique interactions? \(30!\) But we have a problem here, for every finger(let's say for 3rd finger).. \(G_3\) has to come before \(S_3\), and \(S_3\) has to come before \(B_3\) So this has to be our order \(G_3\)\(S_3\)\(B_3\) And this has to be true for all the fingers. What I mean to say is, for any of the \(30!\) arrangements, and for every finger, \(G_n\)\(S_n\)\(B_n\) has to be the order of wearing rings. These interactions may be separated by any number of interactions of other fingers, but this has to be the order. We know that in any of the \(30!\) arrangements, there will be interactions for each finger arranged in some order separated by any number of interactions.. for example.. \(G_n\)\(S_n\)\(B_n\) \(S_n\)\(G_n\)\(B_n\) \(B_n\)\(S_n\)\(G_n\)... \(3!\) = 6 ways. But we have to count only one of these orders \(G_n\)\(S_n\)\(B_n\) This has to be true for all the 10 fingers. So we have to divide \(30!\) by \(6\) for each finger to remove over counting. Final Answer \(\frac{30!}{6^{10}}\) E. mikemcgarry Bunuelcan something of this sort be expected on the GMAT? The question is tagged as Out of scope/Too hard, so no. Also, as Mike mentioned above, the wording is not as precise as GMAT uses.
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Re: Zeynep (the lady of the rings) has 10 identical gold, 10 identical sil
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10 Jul 2017, 10:43
Quote: leve wrote: Zeynep (the lady of the rings) has 10 identical gold, 10 identical silver and 10 identical bronze rings to wear in a particular ceremony. She will wear exactly one gold, one silver and one bronze (total of 3) rings on each of her 10 fingers such that for each finger she will wear one gold ring at the bottom, closest to the knuckle; then one silver ring; then one bronze ring in the most distal position. (According to the rituals of the ceremony this exact arrangement is required on each of her ten fingers.) In how many ways can she execute the processing of putting these 30 rings on her 10 (distinguishable) fingers, so that she ends up with this fixed outcome?
(Example of a possible order: a gold ring to the left index finger, then another gold ring to the right little finger, then another gold ring to the right middle finger, then a silver ring to left index finger, then a bronze ring to left index finger... and so on.)
A) \(\frac{(10!)^3}{2^{10}}\)
B) \(\frac{(10!)^3}{3^{10}}\)
C) \(\frac{30!}{2^{10}}\)
D) \(\frac{30!}{3^{10}}\)
E) \(\frac{30!}{6^{10}}\)
Source: ITU GREGMAT prep seminar. This question was presented as a possible very hard (top level) GREGMAT question. It's tricky but can be done under 2 minutes. Here's my answer: E Note that to wear 30 rings she needs to choose her fingers 30 times (once for each ring)  each finger thrice (once for G, once for S and once for B). Now, she has, F1, F2, F3,.., F10 fingers. One possibility is that the lady wears GSB in F1, then GSB on F2, ... , then GSB on F10 This can be written as, {F1, F1, F1}, {F2, F2, F2},..., {F10, F10, F10}. Another possibility is: All Grings first, from F1F10, then all S from F1F10, then all B from F1F10 > {F1, F2,.., F10}, {F1, F2,..., F10}, {F1, F2,.., F10}. And so on... So all we need to calculate is all possible arrangements of 3 F1's, 3 F2's,... and 3 F10's (Total 3x10 = 30 objects). Hence, \(\frac{30!}{(3!)^{10}} = \frac{30!}{6^{10}}\)




Re: Zeynep (the lady of the rings) has 10 identical gold, 10 identical sil
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