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Post subject: A certain stock exchange designates each stock with a one, t  Posted: Sat Oct 18, 2008 1:06 pm |
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A certain stock exchange designates each stock with a one, two or three 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
OA is E.....18,278.
Please explain this question?
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scrooge
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Post subject: IMO E Posted: Sun Oct 19, 2008 12:00 pm |
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since repeated letters is ok we take permutation : 1 letter code can be obtained in 26 ways
2 letter can be 26 * 26 ways
3 letter can 26 * 26 * 26
26 + 26^2+26^3 IMO E
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Guest
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Post subject: Posted: Sun Nov 02, 2008 8:48 pm |
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Is there a way to solve this problem using permutations? For example, if we are trying to find the # of 2 letter codes using 26 letters can you say:
26! / 2! 24! --> 26 * 25 / 2 = 650.
This is obviously wrong since 26^2 = 675.
What am I doing wrong?
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RonPurewal
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Post subject: Posted: Fri Nov 14, 2008 6:14 am |
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| ManhattanGMAT Staff |
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Posts: 7146
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Anonymous wrote: Is there a way to solve this problem using permutations? For example, if we are trying to find the # of 2 letter codes using 26 letters can you say:
26! / 2! 24! --> 26 * 25 / 2 = 650.
This is obviously wrong since 26^2 = 675.
What am I doing wrong?
well, this problem doesn't really involve permutations. in a bona fide permutation, you're not allowed to re-use the letters, whereas this problem permits the re-use of letters.
for example, there's nothing wrong with the stock code LLL, which would be disallowed in an actual permutation.
by the way, 26 x 26 is 676, not 675. i'll assume you just mis-typed that, since two even numbers clearly won't give a product like that.
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RonPurewal
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Post subject: Re: IMO E Posted: Fri Nov 14, 2008 6:16 am |
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| ManhattanGMAT Staff |
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Posts: 7146
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scrooge wrote: since repeated letters is ok we take permutation : 1 letter code can be obtained in 26 ways
2 letter can be 26 * 26 ways
3 letter can 26 * 26 * 26
26 + 26^2+26^3 IMO E
correct.
by the way, note that there's no need to actually calculate the values of 26^2 and 26^3. instead, just notice that the units digits of the answer choices are all different, no doubt by design. since this is the case, you can just deal with the units digits only, rather than wasting time multiplying out 676 x 26 to find the perfect cube.
no matter how many 6's you multiply together, you still get a number that ends with a 6. therefore, you're basically just adding together three 6's, which gives 18 --> the number will end with an 8.
answer = (e).
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rahul_bitsp
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Post subject: Re: A certain stock exchange designates each stock with a one, t Posted: Mon Aug 24, 2009 1:54 pm |
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Hello Ron.
The question mentions that
"letters may be repeated and if the same letters used in a different order constitute a different code"
but I fail to understand a small issue.
In a 2 code scenario, when we multiple 26 * 26 = 676, are we not including the cases where the first and second symbols are same? I mean don't we count AA and AA (BB or BB) twice ?
I hope you got my question.
Rgds,
Rahul.
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RonPurewal
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Post subject: Re: A certain stock exchange designates each stock with a one, t Posted: Fri Sep 25, 2009 10:51 pm |
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| ManhattanGMAT Staff |
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rahul_bitsp wrote: In a 2 code scenario, when we multiple 26 * 26 = 676, are we not including the cases where the first and second symbols are same? I mean don't we count AA and AA (BB or BB) twice ? nope. each distinct "two-letter word" is counted exactly once. in this approach, there's no difference between "AA", which is counted only once, and, say, "DJ", which is also counted only once. you are correct in that "DJ" and "JD" are counted separately, but that's because they're different "words". try it yourself, with just A's and B's. the multiplication gives 2 x 2 = 4. list them yourself: AA, AB, BA, BB. total 4. the multiplication works.
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sudaif
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Post subject: Re: A certain stock exchange designates each stock with a one, t Posted: Tue Aug 03, 2010 1:37 pm |
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Posts: 126
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if the ORDER DID NOT MATTER, then with the slot method
would it become:
26/1! + 26*26/2! + 26*26*26/3!
?
Thanks
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RonPurewal
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Post subject: Re: A certain stock exchange designates each stock with a one, t Posted: Thu Aug 05, 2010 9:38 am |
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| ManhattanGMAT Staff |
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Posts: 7146
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sudaif wrote: if the ORDER DID NOT MATTER, then with the slot method
would it become:
26/1! + 26*26/2! + 26*26*26/3!
?
Thanks there's no simple way to do "order doesn't matter" if you allow repetition. in situations where repetition is allowed, there is actually no known formula for combinations in which order doesn't matter. so if you get a problem like that, your only recourse will be to make a list, and count the possibilities. it goes without saying that they would only do that with very limited possibilities (i.e., nowhere near the number of outcomes that would happen here).
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pad_bathuu
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Post subject: Re: A certain stock exchange designates each stock with a one, t Posted: Thu Jun 16, 2011 6:22 am |
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I think solution to this problem is...
1.First set of lettters A1,B1, C1.... etc.
Then we have 26 stocks(different stock codes).
2.Second set of letters A2, B2, C2... etc.
Now we have 26+26=52 different letters.
2 letter code out of 52 letters (order matters)
52!/(2!*50!)=1326(different stock codes).
3.Third set of letters A3, B3, C3..... etc
Now we have 26+26+26=78 different letters.
3 letter code out of 78 letters (order matters)
78!/(3!*75!)=76076
The number of total codes for the stock exchange is 76076+1326+26=77428.
This is what I think, but GMAT answer is E.
Please let me know better way to figure out the problem.
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RonPurewal
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Post subject: Re: A certain stock exchange designates each stock with a one, t Posted: Fri Jun 24, 2011 6:11 am |
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| ManhattanGMAT Staff |
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Posts: 7146
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pad_bathuu wrote: Please let me know better way to figure out the problem. did you read the thread? the "better way" you're seeking is right here in the thread.
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