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A certain stock echange designates each stock with a [#permalink]
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A certain stock exchange designates each stock with a 1, 2, or 3 letter code, where each letter is selected from the 26 letters of the alphabet. If the letters may be repeated and if the same letters used in a different order constitute a different code, how many different stocks is it possible to uniquely designate with these codes? A. 2,951 B. 8,125 C. 15,600 D. 16,302 E. 18,278
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Re: A certain stock echange designates each stock with a [#permalink]
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chicagocubsrule wrote: A certain stock exchange designates each stock with a 1, 2, or 3 letter code, where each letter is selected from the 26 letters of the alphabet. If the letters may be repeated and if the same letters used in a different order constitute a different code, how many different stocks is it possible to uniquely designate with these codes?
a) 2,951 b) 8,125 c) 15,600 d) 16,302 e) 18,278 Answer e. if each letter is the same: 26 different combinations 2 letters the same 26^2 all different 26^3 26^3 + 26^2 + 26 = 18278



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chicagocubsrule wrote: A certain stock exchange designates each stock with a 1, 2, or 3 letter code, where each letter is selected from the 26 letters of the alphabet. If the letters may be repeated and if the same letters used in a different order constitute a different code, how many different stocks is it possible to uniquely designate with these codes?
a) 2,951 b) 8,125 c) 15,600 d) 16,302 e) 18,278 1 letter codes = 26 2 letter codes = 26^2 3 letter codes = 26^3 Total = 26 + 26^2 + 26^3 The problem we are faced now is how to get the answer quickly. Note that the units digit of 26+26^2+26^3 would be (6+6+6=18) 8. Only one answer choice has 8 as unit digit: E (18,278). So I believe, even not calculating 26+26^2+26^3, that answer is E.
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Re: A certain stock echange designates each stock with a [#permalink]
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07 Feb 2010, 12:27
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The important language in this one is "letters may be repeated" (i.e. selection with replacement) and "letters used in a different order constitute a different code" (i.e. order doesn't matter). There are 26 onedigit codes. There are 26*26 twodigit codes. There are 26*26*26 threedigit codes. (Note: not 26*25*24, because letters can repeat) There several ways to do the actual calculation, but I did it this way: 26 + 26*26 + 26*26*26 26 (1 + 26 + 26^2) 26 (27 + 676) 26 (703) 18278 Answer (D) 16302 seems to correspond to the following: 26 + 26*26 + 26*25*24. Answer (C) 15600 is 26+ 26^2 less than (D), so it corresponds to: 26*25*24. Answer (B) 8125 is 25*25*13, which is suppose could be arrived at by doing 25*25*26/2. Answer (A) 2951 is 13*227, and I'm not sure what setup error would lead one to arrive at this. The wrong answers are interesting to examine, because they reveal what errors the GMAT writers suspect people will make.
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08 Feb 2010, 04:00
Thanks. I was doing 26+26.25+26.25.24 That would be correct if the letters could not be repeated, wouldnt be? Regards,
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12 Feb 2010, 18:11
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Yes. Your answer will be correct in that way.
Also, I was just thinking since the last digit of the options are all different, we can just calculate the last digit for this instead of doing the actual multiplication and additions: 26 last digit: 6 26^2 last digit: 6 26^3 last digit: 6 Sum = 6+6+6 = x8  so answer should end in 8  option E



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22 Feb 2010, 03:21
esledge wrote: Great shortcut! @esledge In this question, don't you think that order is important. AB is a different code than BA. Therefore, shouldn't we use permutations instead of combinations? I was expecting something like 26C1.26C1.2! for the two digit codes & 26c1.26C1.26C1.3! for 3 digit codes! Please correct me!



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22 Feb 2010, 14:10
honeyrai wrote: In this question, don't you think that order is important. AB is a different code than BA. Therefore, shouldn't we use permutations instead of combinations?
I was expecting something like 26C1.26C1.2! for the two digit codes & 26c1.26C1.26C1.3! for 3 digit codes!
Please correct me! Let's look at the two digit codes: A and B are both among the 26 letters from which you select the first digit: 26C1. A and B are both among the 26 letters from which you select the second digit: 26C1. Thus, with (26C1)(26C1) you are already including AB and BA (and AA and BB, etc.), so no need to increase the count with some multiplier for "shuffling."
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01 Nov 2010, 19:19
The number of combinations for a stock w/ one letter is simply 26 (26 letters).
The number of combinations for a stock w/ two letters is 26*26 = 676.
The number of combinations for a stock w/ three letters is 26*26*26 = 17576.
Summing all of the possible combinations results in 17576 + 676 + 26 = 18278, hence answer E.



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31 Jan 2011, 09:48
Bunuel wrote: chicagocubsrule wrote: The problem we are faced now is how to get the answer quickly. Note that the units digit of 26+26^2+26^3 would be (6+6+6=18) 8. Only one answer choice has 8 as unit digit: E (18,278). So I believe, even not calculating 26+26^2+26^3, that answer is E. The OA is E. Thanks for poininting out how to spot the correct answer  it took me miserable 4 minutes to multiply 26*26*26 and still I made a wrong calculation



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14 Feb 2013, 09:18
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Bunuel wrote: chicagocubsrule wrote: A certain stock exchange designates each stock with a 1, 2, or 3 letter code, where each letter is selected from the 26 letters of the alphabet. If the letters may be repeated and if the same letters used in a different order constitute a different code, how many different stocks is it possible to uniquely designate with these codes?
a) 2,951 b) 8,125 c) 15,600 d) 16,302 e) 18,278 1 letter code=26 2 letter code=26^2 3 letter code=26^3 Total=26+26^2+26^3 The problem we are faced now is how to get the answer quickly. Note that the units digit of 26+26^2+26^3 would be (6+6+6=18) 8. Only one answer choice has 8 as unit digit: E (18,278). So I believe, even not calculating 26+26^2+26^3, that answer is E. Hi Bunuel, Firstly let me say that i fully understand your explanation and it makes perfect sense. I am however, finding it difficult to understand why we can't plug in the numbers into the permutations formula i.e. 26+Pm26,2 + Pm26,3 =16,276 which is well short of the 18,278 answer. I'm just wondering when to apply the approach you mentioned above and when to apply the Permutations formula. Thanks!



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14 Feb 2013, 10:25
26 + 26^2 + 26^3 = 26+676+17576=18278



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iwillbeatthegmat wrote: Bunuel wrote: chicagocubsrule wrote: A certain stock exchange designates each stock with a 1, 2, or 3 letter code, where each letter is selected from the 26 letters of the alphabet. If the letters may be repeated and if the same letters used in a different order constitute a different code, how many different stocks is it possible to uniquely designate with these codes?
a) 2,951 b) 8,125 c) 15,600 d) 16,302 e) 18,278 1 letter code=26 2 letter code=26^2 3 letter code=26^3 Total=26+26^2+26^3 The problem we are faced now is how to get the answer quickly. Note that the units digit of 26+26^2+26^3 would be (6+6+6=18) 8. Only one answer choice has 8 as unit digit: E (18,278). So I believe, even not calculating 26+26^2+26^3, that answer is E. Hi Bunuel, Firstly let me say that i fully understand your explanation and it makes perfect sense. I am however, finding it difficult to understand why we can't plug in the numbers into the permutations formula i.e. 26+Pm26,2 + Pm26,3 =16,276 which is well short of the 18,278 answer. I'm just wondering when to apply the approach you mentioned above and when to apply the Permutations formula. Thanks! Good question. +1. Notice that we are told that the letters may be repeated, so AA, BBB, ACC, CAA, .... codes are possible. Now, 26P2 is the number of ways we can choose 2 distinct letters out of 26 when the order matters, thus it doesn't account for the cases like AA, AAA, ABB, ... Hope it's clear.
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15 Feb 2013, 04:04
Perfectly clear! The repetition disqualifies the permutations formula.
Thanks alot Bunuel!



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23 Mar 2013, 11:12
Also note that you do not have to multiply everything out  just look at the UNITS DIGIT for each number that you multiply/add
26*26*26 = ONES DIGIT IS 6 26*26 = ONES DIGIT IS 6 26 = ONES DIGIT IS 6
The ones digit of the final answer will be 6 + 6 + 6 ... which is 18
The only answer with an 8 in the ones digit is E



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18 Sep 2013, 03:56
lagomez wrote: chicagocubsrule wrote: A certain stock exchange designates each stock with a 1, 2, or 3 letter code, where each letter is selected from the 26 letters of the alphabet. If the letters may be repeated and if the same letters used in a different order constitute a different code, how many different stocks is it possible to uniquely designate with these codes?
a) 2,951 b) 8,125 c) 15,600 d) 16,302 e) 18,278 Answer e. if each letter is the same: 26 different combinations 2 letters the same 26^2 all different 26^3 26^3 + 26^2 + 26 = 18278 what does this statement exactly mean "if the same letters used in a different order constitute a different code"
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honchos wrote: lagomez wrote: chicagocubsrule wrote: A certain stock exchange designates each stock with a 1, 2, or 3 letter code, where each letter is selected from the 26 letters of the alphabet. If the letters may be repeated and if the same letters used in a different order constitute a different code, how many different stocks is it possible to uniquely designate with these codes?
a) 2,951 b) 8,125 c) 15,600 d) 16,302 e) 18,278 Answer e. if each letter is the same: 26 different combinations 2 letters the same 26^2 all different 26^3 26^3 + 26^2 + 26 = 18278 what does this statement exactly mean "if the same letters used in a different order constitute a different code"It means that the order of the letters matters. For example, code AB is different from BA. Similar questions to practice: allofthestocksontheoverthecountermarketare126630.htmlifacodewordisdefinedtobeasequenceofdifferent126652.htmlthesimplasticlanguagehasonly2uniquevaluesand105845.htmla4lettercodewordconsistsoflettersabandcifthe59065.htmla5digitcodeconsistsofonenumberdigitchosenfrom132263.htmlacompanythatshipsboxestoatotalof12distribution95946.htmlacompanyplanstoassignidentificationnumberstoitsempl69248.htmlthesecuritygateatastoragefacilityrequiresafive109932.htmlallofthebondsonacertainexchangearedesignatedbya150820.htmlalocalbankthathas15branchesusesatwodigitcodeto98109.htmlaresearcherplanstoidentifyeachparticipantinacertain134584.htmlbakersdozen12878220.html#p1057502inacertainappliancestoreeachmodeloftelevisionis136646.htmlm04q29colorcoding70074.htmljohnhas12clientsandhewantstousecolorcodingtoiden107307.htmlhowmany4digitevennumbersdonotuseanydigitmorethan101874.htmlacertainstockexchangedesignateseachstockwitha85831.html
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18 Jan 2014, 06:48
iwillbeatthegmat wrote: Bunuel wrote: chicagocubsrule wrote: A certain stock exchange designates each stock with a 1, 2, or 3 letter code, where each letter is selected from the 26 letters of the alphabet. If the letters may be repeated and if the same letters used in a different order constitute a different code, how many different stocks is it possible to uniquely designate with these codes?
a) 2,951 b) 8,125 c) 15,600 d) 16,302 e) 18,278 1 letter code=26 2 letter code=26^2 3 letter code=26^3 Total=26+26^2+26^3 The problem we are faced now is how to get the answer quickly. Note that the units digit of 26+26^2+26^3 would be (6+6+6=18) 8. Only one answer choice has 8 as unit digit: E (18,278). So I believe, even not calculating 26+26^2+26^3, that answer is E. Hi Bunuel, Firstly let me say that i fully understand your explanation and it makes perfect sense. I am however, finding it difficult to understand why we can't plug in the numbers into the permutations formula i.e. 26+Pm26,2 + Pm26,3 =16,276 which is well short of the 18,278 answer. I'm just wondering when to apply the approach you mentioned above and when to apply the Permutations formula. Thanks! 1 letter code: 26 2letter code: P(26,2) + 26 {P(26,2): 2 different numbers and different orders; 26: 2 same numbers} 3letter code: P(26,3) + P(26, 2)C(3, 1) + 26 {P(26,3): 3 different numbers and different orders; P(26, 2)C(3, 1): 2 different numbers, one of which repeats; 26: 3 same numbers} Hope it helps to understand.



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29 Oct 2014, 01:59
I dont know why , but I was thinking for one letter, it's 26, Then for 2 same ones it would be 26^2 2 different ones would mean 26*25 * 2 (because a different order) 3 same would be 26^3, and 3 different would be 26*25*24*3!....Where am I (obviously) double counting?



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29 Oct 2014, 03:48
usre123 wrote: I dont know why , but I was thinking for one letter, it's 26, Then for 2 same ones it would be 26^2 2 different ones would mean 26*25 * 2 (because a different order) 3 same would be 26^3, and 3 different would be 26*25*24*3!....Where am I (obviously) double counting? How is 26^2 the number of two same letter words? How is 26^3 the number of three same letter words? Isn't both 26? AA, BB, CC, ..., ZZ and AAA, BBB, CCC, DDD, ..., ZZZ? 26^2 gives the number of ALL 2letter words possible, the same way as 26^3 gives the number of ALL 3letter words possible.
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