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Amy is organizing her bookshelves and finds that she has
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Updated on: 12 Apr 2014, 07:54
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Amy is organizing her bookshelves and finds that she has 10 different types of books. She then codes each book with either a single letter or a pair of two different letters. If each type of book is uniquely represented by either a single letter or pair of letters, what is the smallest number of letters Amy will need to create the codes for all 10 types of books? (Assume the order of letters in a pair does not matter.) A) 3 B) 4 C) 5 D) 10 E) 20 OE: The question asks for the smallest value of n, such that (n + nC2) = 10 (n represents the number of letters. In this equation, n by itself is for singleletter codes and nC2 is for twoletter codes).
At this point, you'd need to pick numbers, since there's really no easy way to solve nC2 = (10 – n) without a calculator. Looking at the answer choices, you can eliminate 10 and 20, so you can quickly narrow down the values you need to test. (i.e. (10 – n) suggests n can not be greater than 10.)
As a general rule, whenever you're asked for the smallest value that satisfies a condition, start by testing the smallest number in the answers. Conversely, if you're asked for the largest value, start with the greatest answer. Plugin n=4 to (n + nC2) = (4 + 4C2) = 4 + (4x3 /2) = (4 + 6) = 10 Hi, I want to know whether this reverse combination calculation is the GMAT style questions, please. I have not seen one in the Official question.
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Originally posted by goodyear2013 on 11 Apr 2014, 15:34.
Last edited by goodyear2013 on 12 Apr 2014, 07:54, edited 1 time in total.



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Re: Amy is organizing her bookshelves and finds that she has
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11 Apr 2014, 19:11
I wrote all possible answer choices down: A AB (since Order does not matter we will not take BA into account) B CA CB C DA DB DC D



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Re: Amy is organizing her bookshelves and finds that she has
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11 Apr 2014, 23:49
Fairly straightforward question. No reason for it to not be "GMAT style" a) Three letters give : 3 + 3C2 = 3 + 3 = 6 b) Four letters give : 4 + 4C2 = 4 + 12 = 16 So, answer is B
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Re: Amy is organizing her bookshelves and finds that she has
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12 Apr 2014, 00:45
MacFauz wrote: Fairly straightforward question. No reason for it to not be "GMAT style"
a) Three letters give : 3 + 3P2 = 3 + 6 = 9 So, we will need more than 3 So, answer is B Hello  Just a quick thing, the question says "the order of the letter does not matter", so we should be using combinations and permutations.  Kudos, if my post helped



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Re: Amy is organizing her bookshelves and finds that she has
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12 Apr 2014, 01:09
Oops.. Sorry.. My mistake.. Edited above post.. Thanks Posted from my mobile device
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Re: Amy is organizing her bookshelves and finds that she has
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12 Apr 2014, 03:26
goodyear2013 wrote: Amy is organizing her bookshelves and finds that she has 10 different types of books. She then codes each book with either a single letter or a pair of two different letters. If each type of book is uniquely represented by either a single letter or pair of letters, what is the smallest number of letters Amy will need to create the codes for all 10 types of books? (Assume the order of letters in a pair does not matter.) A) 3 B) 4 C) 5 D) 10 E) 20 OE: The question asks for the smallest value of n, such that (n + nC2) = 10 (n represents the number of letters. In this equation, n by itself is for singleletter codes and nC2 is for twoletter codes).
At this point, you'd need to pick numbers, since there's really no easy way to solve nC2 = (10 – n) without a calculator. Looking at the answer choices, you can eliminate 10 and 20, so you can quickly narrow down the values you need to test. (i.e. (10 – n) suggests n can not be less than 10.)
As a general rule, whenever you're asked for the smallest value that satisfies a condition, start by testing the smallest number in the answers. Conversely, if you're asked for the largest value, start with the greatest answer. Plugin n=4 to (n + nC2) = (4 + 4C2) = 4 + (4x3 /2) = (4 + 6) = 10 Hi, I want to know whether this reverse combination calculation is the GMAT style questions, please. I have not seen one in the Official question. Similar questions to practice:eachstudentatacertainuniversityisgivenafourcharact151945.htmlallofthestocksontheoverthecountermarketare126630.htmlifacodewordisdefinedtobeasequenceofdifferent126652.htmla4lettercodewordconsistsoflettersabandcifthe59065.htmla5digitcodeconsistsofonenumberdigitchosenfrom132263.htmlacompanythatshipsboxestoatotalof12distribution95946.htmlacompanyplanstoassignidentificationnumberstoitsempl69248.htmlthesecuritygateatastoragefacilityrequiresafive109932.htmlallofthebondsonacertainexchangearedesignatedbya150820.htmlalocalbankthathas15branchesusesatwodigitcodeto98109.htmlaresearcherplanstoidentifyeachparticipantinacertain134584.htmlbakersdozen12878220.html#p1057502inacertainappliancestoreeachmodeloftelevisionis136646.htmlm04q29colorcoding70074.htmljohnhas12clientsandhewantstousecolorcodingtoiden107307.htmlhowmany4digitevennumbersdonotuseanydigitmorethan101874.htmlacertainstockexchangedesignateseachstockwitha85831.htmlthesimplasticlanguagehasonly2uniquevaluesand105845.htmlm04q29colorcoding70074.htmlaresearcherplanstoidentifyeachparticipantinacertain13458420.htmlHope this helps.
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Amy is organizing her bookshelves and finds that she has
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24 Apr 2017, 07:26
This looks like a pattern question. i tried to just understand what they were asking and picked out some letters to test: if 1 letter, only allowed 1 code (i.e. A) if 2 letters, allowed 3 codes (i.e. A, B, AB) if 3 letters, allowed 6 codes (i.e. A, B, C, AB, AC, BC) if 4 letters, allowed 10 codes (i.e. A, B, C, D, AB, AC, AD, BC, BD, CD) kudos if helpful



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Amy is organizing her bookshelves and finds that she has
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24 Apr 2017, 07:32
three letters could be enough for uniquely naming 9 books i.e (3! *3) and remaining 1 could be named by taking one more letter under consideration. Least total letters would be 10



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Amy is organizing her bookshelves and finds that she has 10 different
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Amy is organizing her bookshelves and finds that she has 10 different types of books. She then codes each book with either a single letter or a pair of two different letters. If each type of book is uniquely represented by either a single letter or pair of letters, what is the smallest number of letters Amy will need to create the codes for all 10 types of books? (Assume the order of letters in a pair does not matter.) A) 3 B) 4 C) 5 F) 10 E) 20 Questions: 1) How do i rephrase this question? The Kaplan answer is not so easy to understand ( at least for me) 2) without getting into too many calculations, how can i solve this? Thank you for your help.
Originally posted by Cinematiccuisine on 04 Sep 2018, 16:46.
Last edited by chetan2u on 04 Sep 2018, 20:11, edited 1 time in total.
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Re: Amy is organizing her bookshelves and finds that she has 10 different
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04 Sep 2018, 17:42
Cinematiccuisine wrote: Amy is organizing her bookshelves and finds that she has 10 different types of books. She then codes each book with either a single letter or a pair of two different letters. If each type of book is uniquely represented by either a single letter or pair of letters, what is the smallest number of letters Amy will need to create the codes for all 10 types of books? (Assume the order of letters in a pair does not matter.) A) 3 B) 4 C) 5 F) 10 E) 20 Questions: 1) How do i rephrase this question? The Kaplan answer is not so easy to understand ( at least for me) 2) without getting into too many calculations, how can i solve this? Thank you for your help. Hi Cinematiccuisine! Happy to help Since the numbers we're dealing with here are small, it is probably easiest to just solve this manually, at first, at least to understand what the question is asking about. Amy is labeling each type of book with either one or two letters. If she is using just one letter, this means there is only one code she can use: A If she is using two letters, then there are three: A B AB We are told that the order of the letters doesn't matter, so we don't have to worry about counting AB and BA separately. Now, with three letters, we have all of the above combinations, plus three more: C AC BC That's six total. If we add one more, then we have those six plus four more: D AD BD CD And that give us 10 possible codes, which is what the question is asking for So the answer is 4. Now, if the number was much bigger, then we might need to create a formula to solve this. The easiest way to do that is to just notice the pattern. Every time we add a new letter, we get one more code (just that letter), plus the number of letters that we already have, which create the combinations of mixed letters (like AD, BD, CD). So adding C gave us 3 more codes, adding D gives us 4 more codes, adding E gives us 5 more codes, and so on. We can use that pattern to determine the number of possible codes at pretty much any step We can write this out in terms of combinations, too. If N is the number of letters that we're using, then the number of codes can be written as: N + NC2 So with four letters, we get 4 + 4C2 = 10 different codes. I hope that helps! Carolyn
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Amy is organizing her bookshelves and finds that she has 10 different
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04 Sep 2018, 20:20
Cinematiccuisine wrote: Amy is organizing her bookshelves and finds that she has 10 different types of books. She then codes each book with either a single letter or a pair of two different letters. If each type of book is uniquely represented by either a single letter or pair of letters, what is the smallest number of letters Amy will need to create the codes for all 10 types of books? (Assume the order of letters in a pair does not matter.) A) 3 B) 4 C) 5 F) 10 E) 20 Questions: 1) How do i rephrase this question? The Kaplan answer is not so easy to understand ( at least for me) 2) without getting into too many calculations, how can i solve this? Thank you for your help. Please put the first few words as the topic name. What it means? 10 books have to be given a code as a single letter such as A, B etc or as a double letter such as AB, BA, AC etc. AB and BA will be different as both are different as code SolutionLet the number of letters required is n, So single digits will be n Two digits will be nC2........ .we are not multiplying with 2! because order does not matter. So \(n+nC2\geq{10}\)........\(n+\frac{n!}{(n2)!2!}\geq{10}........n+n(n1)/2\geq{10}...............(2n+n^2n}\geq{20}............n(n+1)\geq{20}.........n(n+1)\geq{4*5}..\) So minimum value = 4 B
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