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Re: A researcher plans to identify each participant in a certain medical
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25 Feb 2014, 08:39
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 medical
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13 Mar 2014, 14:03
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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Re: A researcher plans to identify each participant in a certain medical
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14 Mar 2014, 01:48
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 medical
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06 Apr 2014, 11:38
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 medical
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06 Apr 2014, 12:09
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 medical
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02 Sep 2014, 17:53
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 medical
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02 Sep 2014, 19:16
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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Re: A researcher plans to identify each participant in a certain medical
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10 Sep 2014, 10:41
Hi Bunuel,
I'm confused on when you show that n! = (n2)!*(n1)*n...why is n! only limited to those 3 factors? I guess the question is why do you start at (n2)!?
Thanks.!
\(C^2_n=\frac{n!}{2!(n2)!}\). 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 medical
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10 Sep 2014, 10:54
kevn1115 wrote: Hi Bunuel,
I'm confused on when you show that n! = (n2)!*(n1)*n...why is n! only limited to those 3 factors? I guess the question is why do you start at (n2)!?
Thanks.!
\(C^2_n=\frac{n!}{2!(n2)!}\). 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. n! is the product of positive integers from 1 to n, inclusive: n! = 1*2*...*(n4)*(n3)(n2)(n1)n. To simplify \(\frac{n!}{2!(n2)!}\) I wrote n! as (n2)!*(n1)*n this enables us to reduce by (n2)! to get \(\frac{(n1)n}{2}\). Hope it's clear.
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Re: A researcher plans to identify each participant in a certain medical
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10 Sep 2014, 11:06
Thanks Bunuel. I'm sorry, but I'm still not following.
Why couldn't you simplify n! as (n4)!*(n3)!*(n2)!*(n1)!*n? What tells you that n! starts with (n2)! and not a smaller number like (n3)! or further?
What is the significance about starting with "2" using (n2)! and working your way down to "n" when simplifying this?
Thanks.



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Re: A researcher plans to identify each participant in a certain medical
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10 Sep 2014, 11:21
kevn1115 wrote: Thanks Bunuel. I'm sorry, but I'm still not following.
Why couldn't you simplify n! as (n4)!*(n3)!*(n2)!*(n1)!*n? What tells you that n! starts with (n2)! and not a smaller number like (n3)! or further?
What is the significance about starting with "2" using (n2)! and working your way down to "n" when simplifying this?
Thanks. I think you need to brush up fundamentals. The factorial of a nonnegative integer n, denoted by \(n!\), is the product of all positive integers less than or equal to n. For example: \(4!=1*2*3*4=24\). Now, consider this: \(10! = 1*2*3*...*8*9*10\) can also be written as \(9!*10 = (1*2*3*...*8*9)*10 = 10!\) or \(8!*9*10 = (1*2*3*...*8)*9*10 = 10!\), or \(7!*8*9*10 = (1*2*3*...*7)*8*9*10 = 10!\) ... Similarly \(n!\) (n factorial) can be written as \((n2)!*(n1)*n\) (n2 factorial multiplied by (n1)*n). We need to write it that way in order to be able to reduce by \((n2)!\). Hope it's clear.
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Re: A researcher plans to identify each participant in a certain medical
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27 Dec 2014, 10:22
Hi All, In these sorts of questions, when the answer choices are relatively small, it's often fairly easy to "brute force" the correct answer and avoid complicated calculations entirely. BrainLab's idea to just "map out" the possibilities is a relatively simple, effective approach. Since we're asked for the LEAST number of letters that will give us 12 unique codes, we start with Answer A. If we had 4 letters: A, B, C, D 1letter codes: A, B, C, D 2letter alphabetical codes: AB, AC, AD, BC, BD, CD Total Codes = 4 + 6 = 10 This result is TOO LOW. From here, you know that we just need 2 more codes, so adding 1 more letter would give us those extra codes (and more)...but here's the proof that it happens.... If we had 5 letters: A, B, C, D, E 1letter codes: A, B, C, D, E 2letter alphabetical codes: AB, AC, AD, AE, BC, BD, BE, CD, CE, DE Total Codes = 5 + 10 = 15 codes Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: A researcher plans to identify each participant in a certain medical
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08 Feb 2015, 08:44
Bunuel wrote: kevn1115 wrote: Hi Bunuel,
I'm confused on when you show that n! = (n2)!*(n1)*n...why is n! only limited to those 3 factors? I guess the question is why do you start at (n2)!?
Thanks.!
\(C^2_n=\frac{n!}{2!(n2)!}\). 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. n! is the product of positive integers from 1 to n, inclusive: n! = 1*2*...*(n4)*(n3)(n2)(n1)n. To simplify \(\frac{n!}{2!(n2)!}\) I wrote n! as (n2)!*(n1)*n this enables us to reduce by (n2)! to get \(\frac{(n1)n}{2}\). Hope it's clear. Hi bunuel! Sorry for opening the old topic but my question is why are you and everyone else writing the "c(n, k)" upside down. Have a look at the screenshot of the gmat club math book and the equation everyone is referring to. Thanks!
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Re: A researcher plans to identify each participant in a certain medical
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09 Feb 2015, 00:43
ProblemChild wrote: Bunuel wrote: kevn1115 wrote: Hi Bunuel,
I'm confused on when you show that n! = (n2)!*(n1)*n...why is n! only limited to those 3 factors? I guess the question is why do you start at (n2)!?
Thanks.!
\(C^2_n=\frac{n!}{2!(n2)!}\). 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. n! is the product of positive integers from 1 to n, inclusive: n! = 1*2*...*(n4)*(n3)(n2)(n1)n. To simplify \(\frac{n!}{2!(n2)!}\) I wrote n! as (n2)!*(n1)*n this enables us to reduce by (n2)! to get \(\frac{(n1)n}{2}\). Hope it's clear. Hi bunuel! Sorry for opening the old topic but my question is why are you and everyone else writing the "c(n, k)" upside down. Have a look at the screenshot of the gmat club math book and the equation everyone is referring to. Thanks! \(C^1_3\), \(C^3_1\), 3C1, mean the same thing choosing 1 out of 3, just different ways to write. Can it be choosing 3 out of 1???
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Re: A researcher plans to identify each participant in a certain medical
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28 Apr 2015, 22:43
Hi Folks,
Pair would mean 2 elements.So we will consider single and pairs(with 2 elements) only right?



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Re: A researcher plans to identify each participant in a certain medical
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28 Apr 2015, 23:39
kirtivardhan wrote: Hi Folks,
Pair would mean 2 elements.So we will consider single and pairs(with 2 elements) only right? Hi kirtivardhan, You are right, pair means 2 elements. Hence the solution considers both single letters and pair of letters for formation of the code. Let me know at which point you are having a doubt. Regards Harsh
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Re: A researcher plans to identify each participant in a certain medical
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29 Apr 2015, 00:10
Hi Harsh, If we try to solve it by trial and error then for 5 letters it would be 5 single letters and AB,BC,CD,DE in alphabetical order.that makes the total as 9.Then how is the answer 5?



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Re: A researcher plans to identify each participant in a certain medical
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29 Apr 2015, 01:13
kirtivardhan wrote: Hi Harsh, If we try to solve it by trial and error then for 5 letters it would be 5 single letters and AB,BC,CD,DE in alphabetical order.that makes the total as 9.Then how is the answer 5? Hi kirtivardhan, Writing the letters in alphabetical order does not mean using only consecutive alphabets. For a two alphabet combination where the first alphabet is A, every other alphabet coming after A will have the code in in alphabetical order. So, AB, AC, AD and AE are in alphabetical order. Similarly for the first alphabet as B we will have BC, BD & BE in alphabetical order. Again, for C we would have CD & CE and for D we would have DE in alphabetical order. You missed out on the combination of these alphabets. Another way to understand this is for the alphabet A, there are 4 other letters (i.e. B, C, D, E) which can follow A, so A will have 4 possible combinations in alphabetical order. For alphabet B, there are 3 letters (i.e. C, D, E) which can follow B, so B will have 3 possible combinations in alphabetical order. Similarly for C, we will have 2 possible combinations (i.e. D, E) and 1 possible combination (i.e. E) for D. Total possible two alphabet combinations = 4 + 3 + 2 + 1 = 10 Total possible single alphabet combination = 5 Hence, total possible combinations i.e. distinct codes possible = 10 + 5 = 15 > 12.Try out this exercise for 4 letters and you will observe that the total distinct possible codes is less than 12. Hope it's clear Regards Harsh
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Re: A researcher plans to identify each participant in a certain medical
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21 Nov 2015, 16:55
i understand the concept of the combinatorics, but how for ex AD is in alphabetical order? I thought that we are restricted to use a pair of letters only in alphabetical order, ex: AB, BC, CD, etc. that is my understanding of the alphabetical order.



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21 Nov 2015, 23:13
Hi mvictor, There's a difference between alphabetical order and CONSECUTIVE alphabetical order (in the same way that there's a difference between putting integers in numerical order and dealing with consecutive integers). As an example, when dealing with the letters A, B, C and D there are 6 different pairs of letters that you could put in alphabetical order: AB AC AD BC BD CD GMAT assassins aren't born, they're made, Rich
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