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# permutations :S

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Intern
Joined: 08 Jun 2009
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permutations :S [#permalink]  16 Jun 2009, 07:27
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Difficulty:

(N/A)

Question Stats:

33% (02:46) correct 67% (00:00) wrong based on 12 sessions
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A certain stock exchange designates each stock with one-, two-, or three-letter code, where each letter is selected from the 26 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
Current Student
Joined: 13 May 2008
Posts: 141
Schools: LBS
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Kudos [?]: 24 [0], given: 6

Re: permutations :S [#permalink]  16 Jun 2009, 08:10
Jozu wrote:
A certain stock exchange designates each stock with one-, two-, or three-letter code, where each letter is selected from the 26 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

stock with one-- 26
two-, 26*25= 650
or three-letter code, 26*25*24= 15,600

the total is 16,276 ... your option (d) is 16,302 thats intriguingly 26 more than my answer ... now i am probably doing something wrong but i can't figure it out and if you do, do post here b/c i am sort of curious now.
Current Student
Joined: 03 Aug 2006
Posts: 116
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Schools: Haas School of Business
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Re: permutations :S [#permalink]  16 Jun 2009, 08:18
Given that each letter in each stock code can be repeated we need to factor that in our calculations:

$$\text {No of 1 letter codes} = 26$$

$$\text {No of 2 letter codes} = 26\times 26 = 26^2$$

$$\text {No of 3 letter codes} = 26\times 26\times 26 = 26^3$$

$$\text {Total number of codes} = 26 + 26^2 + 26^3 = 18,278$$

The answer is E.

For the last step there is a faster way to get the answer than doing all the multiplication.

26 has 6 in the units digit.
Also 26^2 should have 6 in the units digit.
Similarly 26^3 should have 6 in the units digit.

If you add the units digits of the three it is 18 that would mean 8 would be in the units digit of the final answer and there is only one answer choice with 8 in the units digit i.e. E

See the following thread for more on last digit of a power.

last-digit-of-a-power-70624.html#p521012
Intern
Joined: 08 Jun 2009
Posts: 33
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Kudos [?]: 17 [0], given: 0

Re: permutations :S [#permalink]  16 Jun 2009, 17:03
OA is E.

I'm a little confused here regarding permutations.

How do we know when to use nPr (order is relevant), nCr (order is not relevant) and your method above? I just can't seem to apply them correctly when I encounter such problems.
Current Student
Joined: 03 Aug 2006
Posts: 116
Location: Next to Google
Schools: Haas School of Business
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Kudos [?]: 174 [0], given: 3

Re: permutations :S [#permalink]  16 Jun 2009, 19:33
This problem is testing the fundamental counting principal.

Here is a link to understand it better with some examples.

Also checkout this thread for more on Permutations and Combinations.

permutations-combinations-help-is-on-the-way-10838.html
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GMAT 1: 750 Q49 V42
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Re: permutations :S [#permalink]  16 Jun 2009, 23:28
Expert's post
And of course the Walker's thread that is stickied in this forum: combinations-permutations-and-probability-references-56486.html
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Re: permutations :S   [#permalink] 16 Jun 2009, 23:28
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# permutations :S

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