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Re: A researcher plans to identify each participant in a certain [#permalink]
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25 Nov 2013, 04:47
Lets take A B C D A B C D AB AC AD BC BD CD ABC BCA CBA It is alphabetical and all letter for a particular codes are different.
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25 Nov 2013, 09:39
honchos wrote: Bunuel wrote: sarb wrote: A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?
A. 4 B. 5 C. 6 D. 7 E. 8 Say there are minimum of \(n\) letters needed, then; The # of single letter codes possible would be \(n\) itself; The # of pair of distinct letters codes possible would be \(C^2_n\) (in alphabetical order); We want \(C^2_n+n\geq{12}\) > \(\frac{n(n1)}{2}+n\geq{12}\) > \(n(n1)+2n\geq{24}\) > \(n(n+1)\geq{24}\) > \(n_{min}=5\). Answer: B. Hope it's clear. we can take 1,2 and 3 like A, B, C AB, BC ABC Why did you ignored possibility of 3 or 4 alphabets taken together, this will give us 4 letters? answered in red



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29 Nov 2013, 20:18
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honchos wrote: Bunuel wrote: sarb wrote: A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?
A. 4 B. 5 C. 6 D. 7 E. 8 Say there are minimum of \(n\) letters needed, then; The # of single letter codes possible would be \(n\) itself; The # of pair of distinct letters codes possible would be \(C^2_n\) (in alphabetical order); We want \(C^2_n+n\geq{12}\) > \(\frac{n(n1)}{2}+n\geq{12}\) > \(n(n1)+2n\geq{24}\) > \(n(n+1)\geq{24}\) > \(n_{min}=5\). Answer: B. Hope it's clear. we can take 1,2 and 3 like A, B, C AB, BC ABC Why did you ignored possibility of 3 or 4 alphabets taken together, this will give us 4 letters? The question specifically points out that the combinations can be a 1 digit letter or a 2 digit letter. I used a simple combination as stated in other answers to find out. 1. A 2. B 3. BA 4. C 5. CA 6. CB 7. D 8. DA 9. DB 10. DC 11. E 12. EA STOP. you get the answer as 5 (ABCDE) Also what i have found is that when writing down the combinations with no repeats, it is easier to start with one letter and keep repeating it until you exhausted all the options. this will eliminate confusion. like you start with C and repeat with CA CB and then with D DA DB DC..



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Re: A researcher plans to identify each participant in a certain [#permalink]
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25 Feb 2014, 08:57
Bunuel wrote: sarb wrote: A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?
A. 4 B. 5 C. 6 D. 7 E. 8 Say there are minimum of \(n\) letters needed, then; The # of single letter codes possible would be \(n\) itself; The # of pair of distinct letters codes possible would be \(C^2_n\) (in alphabetical order); We want \(C^2_n+n\geq{12}\) > \(\frac{n(n1)}{2}+n\geq{12}\) > \(n(n1)+2n\geq{24}\) > \(n(n+1)\geq{24}\) > \(n_{min}=5\). Answer: B. Hope it's clear. I have a questions here: How did we get from \(n(n+1)\geq{24}\) to \(n_{min}=5\)



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Re: A researcher plans to identify each participant in a certain [#permalink]
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25 Feb 2014, 09:39
amz14 wrote: Bunuel wrote: sarb wrote: A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?
A. 4 B. 5 C. 6 D. 7 E. 8 Say there are minimum of \(n\) letters needed, then; The # of single letter codes possible would be \(n\) itself; The # of pair of distinct letters codes possible would be \(C^2_n\) (in alphabetical order); We want \(C^2_n+n\geq{12}\) > \(\frac{n(n1)}{2}+n\geq{12}\) > \(n(n1)+2n\geq{24}\) > \(n(n+1)\geq{24}\) > \(n_{min}=5\). Answer: B. Hope it's clear. I have a questions here: How did we get from \(n(n+1)\geq{24}\) to \(n_{min}=5\) By trial and error: If n=4, then n(n+1)=20<24; If n=5, then n(n+1)=30>24. Hence, \(n_{min}=5\). Try similar questions to practice: aresearcherplanstoidentifyeachparticipantinacertain134584.html#p1296049Hope this helps.
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Re: A researcher plans to identify each participant in a certain [#permalink]
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13 Mar 2014, 15:03
Bunuel wrote: sarb wrote: A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?
A. 4 B. 5 C. 6 D. 7 E. 8 Say there are minimum of \(n\) letters needed, then; The # of single letter codes possible would be \(n\) itself; The # of pair of distinct letters codes possible would be \(C^2_n\) (in alphabetical order); We want \(C^2_n+n\geq{12}\) > \(\frac{n(n1)}{2}+n\geq{12}\) > \(n(n1)+2n\geq{24}\) > \(n(n+1)\geq{24}\) > \(n_{min}=5\). Answer: B. Hope it's clear. I am still having problems with this question. Why do we devide the combinations formula into n(n1)/2? Shouldnt it be 2!/n!(2n)! ?



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RebekaMo wrote: Bunuel wrote: sarb wrote: A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?
A. 4 B. 5 C. 6 D. 7 E. 8 Say there are minimum of \(n\) letters needed, then; The # of single letter codes possible would be \(n\) itself; The # of pair of distinct letters codes possible would be \(C^2_n\) (in alphabetical order); We want \(C^2_n+n\geq{12}\) > \(\frac{n(n1)}{2}+n\geq{12}\) > \(n(n1)+2n\geq{24}\) > \(n(n+1)\geq{24}\) > \(n_{min}=5\). Answer: B. Hope it's clear. I am still having problems with this question. Why do we devide the combinations formula into n(n1)/2? Shouldnt it be 2!/n!(2n)! ? \(C^2_n=\frac{n!}{2!(n2)!}\). Now, notice that \(n!=(n2)!*(n1)*n\), hence \(C^2_n=\frac{n!}{2!(n2)!}=\frac{(n2)!*(n1)*n}{2!(n2)!}=\frac{(n1)n}{2}\). Hope it's clear.
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Re: A researcher plans to identify each participant in a certain [#permalink]
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06 Apr 2014, 12:38
Bunuel wrote: sarb wrote: A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?
A. 4 B. 5 C. 6 D. 7 E. 8 Say there are minimum of \(n\) letters needed, then; The # of single letter codes possible would be \(n\) itself; The # of pair of distinct letters codes possible would be \(C^2_n\) (in alphabetical order); We want \(C^2_n+n\geq{12}\) > \(\frac{n(n1)}{2}+n\geq{12}\) > \(n(n1)+2n\geq{24}\) > \(n(n+1)\geq{24}\) > \(n_{min}=5\). Answer: B. Hope it's clear. Probability and Combinatorics are my weakest subjects by far, so please ignore the rudimentary question. When we say \(C^2_n+n\geq{12}\) that means that we are going to find a combination of 2 letters out of a group of n letters which in turn would yield "x" amount of options. Correct? If so, why are we adding the n following that equation and more importantly, how does that equation yield 5? When I factor it out, i get n(n+1) >= 24. That yields 1 and 0. Why am I so off here?My question would be  what does this formula mean and how do you solve it? \(C^2_n+n\geq{12}\) Also, the question is saying that they need to be in alphabetical order, doesn't that mean that order DOES matter? How does that affect the above equation. P.S: For what it's worth, I've read the Combinatorics and Probability strategy guide( Manhattan Gmat) and understand the content of the guide but these two topics still elude me. I'm open to learning from another venue if helpful? EDIT: Simplifying my question.



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Re: A researcher plans to identify each participant in a certain [#permalink]
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06 Apr 2014, 13:09
russ9 wrote: Bunuel wrote: sarb wrote: A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?
A. 4 B. 5 C. 6 D. 7 E. 8 Say there are minimum of \(n\) letters needed, then; The # of single letter codes possible would be \(n\) itself; The # of pair of distinct letters codes possible would be \(C^2_n\) (in alphabetical order); We want \(C^2_n+n\geq{12}\) > \(\frac{n(n1)}{2}+n\geq{12}\) > \(n(n1)+2n\geq{24}\) > \(n(n+1)\geq{24}\) > \(n_{min}=5\). Answer: B. Hope it's clear. Probability and Combinatorics are my weakest subjects by far, so please ignore the rudimentary question. When we say \(C^2_n+n\geq{12}\) that means that a combination of 2 letters out of a group of n letters should yield "x" amount of options. Correct? If so, why are we adding the n following that equation and more importantly, how does that equation yield 5? When I factor it out, i get n(n+1) >= 24. That yields 1 and 0. Why am I so off here? Also, the question is saying that they need to be in alphabetical order, doesn't that mean that order DOES matter? How does that affect the above equation. P.S: For what it's worth, I've read the Combinatorics and Probability strategy guide and understand the content of the guide but these two topics still elude me. I'm open to learning from another venue if helpful? The first advice would be, and I cannot stress this enough, to read the whole thread and follow the links to similar problems. As for your questions: Why we are adding n. The question says that the code can consists of 1 or 2 letters. Now, if we have n letters how many codes we can make? The # of single letter codes possible would be n itself; The # of pair of distinct letters codes possible would be \(C^2_n\). So, out of n letters we can make \(n+C^2_n\) codes: n oneletter codes and \(C^2_n\) twoletter codes. How the equation yields 5By trial and error: If n=4, then n(n+1)=20<24; If n=5, then n(n+1)=30>24. Hence, \(n_{min}=5\). Notice that we have \(n(n+1)\geq{24}\) NOT \(n(n+1)\geq{0}\). About the alphabetical order.Check here: aresearcherplanstoidentifyeachparticipantinacertain134584.html#p1150091 and here: aresearcherplanstoidentifyeachparticipantinacertain134584.html#p1296053Similar questions to practice:eachstudentatacertainuniversityisgivenafourcharact151945.htmlallofthestocksontheoverthecountermarketare126630.htmlifacodewordisdefinedtobeasequenceofdifferent126652.htmla4lettercodewordconsistsoflettersabandcifthe59065.htmla5digitcodeconsistsofonenumberdigitchosenfrom132263.htmlacompanythatshipsboxestoatotalof12distribution95946.htmlacompanyplanstoassignidentificationnumberstoitsempl69248.htmlthesecuritygateatastoragefacilityrequiresafive109932.htmlallofthebondsonacertainexchangearedesignatedbya150820.htmlalocalbankthathas15branchesusesatwodigitcodeto98109.htmlaresearcherplanstoidentifyeachparticipantinacertain134584.htmlbakersdozen12878220.html#p1057502inacertainappliancestoreeachmodeloftelevisionis136646.htmlm04q29colorcoding70074.htmljohnhas12clientsandhewantstousecolorcodingtoiden107307.htmlhowmany4digitevennumbersdonotuseanydigitmorethan101874.htmlacertainstockexchangedesignateseachstockwitha85831.htmlthesimplasticlanguagehasonly2uniquevaluesand105845.htmlm04q29colorcoding70074.htmlHope this helps.
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Re: A researcher plans to identify each participant in a certain [#permalink]
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06 Apr 2014, 13:48
Hi Bunuel,
Thanks for the clarification. I was having a hard time grasping the equation itself but I followed a link to the mathbook topic and that does a good job of explaining why the equation is the way it is.
What I do question is the arrangement of the letters. I did go to the two links you posted and it's still a little unclear. How does the 2Cn equation know to not double count BA and AB. Wouldn't we have to go to the permutation equation for that?



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15 May 2014, 14:53
Bunuel wrote: sarb wrote: A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?
A. 4 B. 5 C. 6 D. 7 E. 8 Say there are minimum of \(n\) letters needed, then; The # of single letter codes possible would be \(n\) itself; The # of pair of distinct letters codes possible would be \(C^2_n\) (in alphabetical order); We want \(C^2_n+n\geq{12}\) > \(\frac{n(n1)}{2}+n\geq{12}\) > \(n(n1)+2n\geq{24}\) > \(n(n+1)\geq{24}\) > \(n_{min}=5\). Answer: B. Hope it's clear. I just don't understand how we get n(n1)/2 out of nC2... Isn't it n!/(n2)! ? Combinations are the worst part of GMAT for me.



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Re: A researcher plans to identify each participant in a certain [#permalink]
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16 May 2014, 01:43
bytatia wrote: Bunuel wrote: sarb wrote: A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?
A. 4 B. 5 C. 6 D. 7 E. 8 Say there are minimum of \(n\) letters needed, then; The # of single letter codes possible would be \(n\) itself; The # of pair of distinct letters codes possible would be \(C^2_n\) (in alphabetical order); We want \(C^2_n+n\geq{12}\) > \(\frac{n(n1)}{2}+n\geq{12}\) > \(n(n1)+2n\geq{24}\) > \(n(n+1)\geq{24}\) > \(n_{min}=5\). Answer: B. Hope it's clear. I just don't understand how we get n(n1)/2 out of nC2... Isn't it n!/(n2)! ? Combinations are the worst part of GMAT for me. Please read the whole thread. A lot of useful staff there. Your question is answered here: aresearcherplanstoidentifyeachparticipantinacertain13458420.html#p1344122Hope it helps.
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Re: A researcher plans to identify each participant in a certain [#permalink]
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18 May 2014, 00:24
Bunuel wrote: sarb wrote: A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?
A. 4 B. 5 C. 6 D. 7 E. 8 Say there are minimum of \(n\) letters needed, then; The # of single letter codes possible would be \(n\) itself; The # of pair of distinct letters codes possible would be \(C^2_n\) (in alphabetical order); We want \(C^2_n+n\geq{12}\) > \(\frac{n(n1)}{2}+n\geq{12}\) > \(n(n1)+2n\geq{24}\) > \(n(n+1)\geq{24}\) > \(n_{min}=5\). Answer: B. Hope it's clear. Every time I see your explanation, problem becomes so easy, but when I tried my own, I hardly get the correct. How to improve my understanding on combination and Probability ?



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18 May 2014, 01:07
gauravsaxena21 wrote: Bunuel wrote: sarb wrote: A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?
A. 4 B. 5 C. 6 D. 7 E. 8 Say there are minimum of \(n\) letters needed, then; The # of single letter codes possible would be \(n\) itself; The # of pair of distinct letters codes possible would be \(C^2_n\) (in alphabetical order); We want \(C^2_n+n\geq{12}\) > \(\frac{n(n1)}{2}+n\geq{12}\) > \(n(n1)+2n\geq{24}\) > \(n(n+1)\geq{24}\) > \(n_{min}=5\). Answer: B. Hope it's clear. Every time I see your explanation, problem becomes so easy, but when I tried my own, I hardly get the correct. How to improve my understanding on combination and Probability ? By studying theory and practicing. Hope this helps.
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13 Aug 2014, 17:49
Bunuel wrote: sarb wrote: A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?
A. 4 B. 5 C. 6 D. 7 E. 8 Say there are minimum of \(n\) letters needed, then; The # of single letter codes possible would be \(n\) itself; The # of pair of distinct letters codes possible would be \(C^2_n\) (in alphabetical order); We want \(C^2_n+n\geq{12}\) > \(\frac{n(n1)}{2}+n\geq{12}\) > \(n(n1)+2n\geq{24}\) > \(n(n+1)\geq{24}\) > \(n_{min}=5\). Answer: B. Hope it's clear. Thanks Bunuel for the explanation. I do need some clarification regarding the C(n,r). How does C(n, 2) = n(n1)/2? Shouldn't it be n! / (2! (n  2)!) Thank you!



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14 Aug 2014, 00:54
ccyang24 wrote: Bunuel wrote: sarb wrote: A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?
A. 4 B. 5 C. 6 D. 7 E. 8 Say there are minimum of \(n\) letters needed, then; The # of single letter codes possible would be \(n\) itself; The # of pair of distinct letters codes possible would be \(C^2_n\) (in alphabetical order); We want \(C^2_n+n\geq{12}\) > \(\frac{n(n1)}{2}+n\geq{12}\) > \(n(n1)+2n\geq{24}\) > \(n(n+1)\geq{24}\) > \(n_{min}=5\). Answer: B. Hope it's clear. Thanks Bunuel for the explanation. I do need some clarification regarding the C(n,r). How does C(n, 2) = n(n1)/2? Shouldn't it be n! / (2! (n  2)!) Thank you! Please read the whole thread. A lot of useful staff there. Your question is answered here: aresearcherplanstoidentifyeachparticipantinacertain13458420.html#p1344122Hope it helps.
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Joined: 10 Jul 2014
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Re: A researcher plans to identify each participant in a certain [#permalink]
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02 Sep 2014, 18:53
Bunuel wrote: sarb wrote: A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?
A. 4 B. 5 C. 6 D. 7 E. 8 Say there are minimum of \(n\) letters needed, then; The # of single letter codes possible would be \(n\) itself; The # of pair of distinct letters codes possible would be \(C^2_n\) (in alphabetical order); We want \(C^2_n+n\geq{12}\) > \(\frac{n(n1)}{2}+n\geq{12}\) > \(n(n1)+2n\geq{24}\) > \(n(n+1)\geq{24}\) > \(n_{min}=5\). Answer: B. Hope it's clear. I don't get how this is B instead of D. Aside from the formula, it says in alphabetical order, so how can you count AC, etc.?



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Re: A researcher plans to identify each participant in a certain [#permalink]
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02 Sep 2014, 20:16
Chin926926 wrote: Bunuel wrote: sarb wrote: A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?
A. 4 B. 5 C. 6 D. 7 E. 8 Say there are minimum of \(n\) letters needed, then; The # of single letter codes possible would be \(n\) itself; The # of pair of distinct letters codes possible would be \(C^2_n\) (in alphabetical order); We want \(C^2_n+n\geq{12}\) > \(\frac{n(n1)}{2}+n\geq{12}\) > \(n(n1)+2n\geq{24}\) > \(n(n+1)\geq{24}\) > \(n_{min}=5\). Answer: B. Hope it's clear. I don't get how this is B instead of D. Aside from the formula, it says in alphabetical order, so how can you count AC, etc.? AC is in alphabetical order while CA is not.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: A researcher plans to identify each participant in a certain
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