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GMAT PREP PROBABILITY?

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GMAT PREP PROBABILITY?

by dferm » Tue Mar 25, 2008 7:55 am
A certain stock exchange designates each stock with a one-, or 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

PLEASE HELP>>
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by tmmyc » Tue Mar 25, 2008 8:53 am
Here is what we know:
-There are three types of codes: one-letter, two-letter, three-letter.
-There are 26 letters to choose from.
-Letters can be repeated.
-Letters in different orders constitute different codes.

Let's calculate each of the three types of codes.

One-letter: 26, since there are 26 to choose from.

Two-letter: 26*26, since the first letter can be any of the 26 letters and so can the second letter. Letters can be repeated (AA) and letters in different orders constitute as difference codes (AC and CA are unique).

Three-letter: 26*26*26 using the same reasoning at the two-letter codes.


To find the total stocks, add all of them up:

(one-letter) + (two-letter) + (three-letter) = (total codes)


Since we don't want to calculate 26^3, we can use some number properties.

Look at the units digit 6. We know that multiplying 6 by itself any number of times will still retain a units digit of 6 (6*6 = 36; 36*6 = 216; the units digit is always 6).

Using this knowledge we can sum the three as follows

(one-letter with units digit 6) + (two-letter with units digit 6) + (three-letter with units digit 6) = (total code with units digit 8)
since 6+6+6=18 and you carry the 1.


The only answer with a units digit of 8 is E.
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by aj5105 » Wed Jul 01, 2009 11:31 pm
Great job, Tim.
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by funx » Tue Jul 28, 2009 6:14 am
Although I understand your answer, I am asking myself why cant one use permutations to solve this problem?

What I mean is the following:

26P3 = 15,600
26P2 = 650
26P1 = 26

This method gives an answer which doesn't match any of the given choices. But I still don't understand why can't you apply permutations here. Can anyone shed some light?

Thanks.
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by kaulnikhil » Tue Jul 28, 2009 7:20 am
funx
u cant use 26 p3 here becauz the question says letters may be repeated had this not been the clause ..ur answer would have been perfectly right .. when u use the above approach u fill the first postion in 26 ways second in 25 and third in 24 ..which is wrong here ..since we can fill all the positions in 26 ways..hope u got it
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