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# GMAT prep combination question

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Manager
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04 Oct 2006, 17:39
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

getting nervous...test time is saturday... Can anyone help explain where my logic wen wrong here?

Question

A certain stock exchange designates each stock with a one , two, or three letter code where each letter is selected from 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) 18278

They way i approached was to use the combo formula

combo (picking 1 out of 26) + Combo (picking 2 out of 26) + combo (picking 3 out of 26)..

I'm guessing my logic doesn't take into consideration the fact that the letters can repeat and that the same letters in a different order constitue a different code..
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04 Oct 2006, 17:48
Here we go.

For the 3 letter case. Number of combinations = 26*26*26, since letters can be repeated

For the 2 leter case 26*26

For 1 letter case 26

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04 Oct 2006, 19:51
rbcola wrote:
Here we go.

For the 3 letter case. Number of combinations = 26*26*26, since letters can be repeated

For the 2 leter case 26*26

For 1 letter case 26

Hey rbcola,
I arrived at E the same way as you, but I have a doubt.

In the two letter combos for example, you cannot distinguish between the two "BB"s. So, there's effectively only one BB. So, you lose 26 two letter combos this way. The same logic for the three letter combos as well.

So, I'm guessing that the answer is something smaller than E, probably D.
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04 Oct 2006, 20:24
I think the answer will be E. To put it more clearly:
(i) Using only 1 letter, there are 26 possibilities
(ii) Using 2 letters:
All letters are different = 26P2 = 650
If letters are repeated (AA, BB) = 26 ways
(iii) Using 3 letters
All letters are different = 26P3 = 15600
All 3 letters are repeated (AAA, BB) = 26 ways
2 letter are repeated (AAB, AAC) = 26*25*3 = 1950

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04 Oct 2006, 23:13
It must be E

Single letter codes---26
Double letter codes --- 26^2
In a double letter code first letter can be anyone of the 26 alphabets and since repetition is allowed second letter also can be anyone of the 26 alphabets i.e. 26x26 = 26^2

Similarly three letter codes ---- 26^3

So 26+26^2+26^3 = 18278
_________________

Last edited by cicerone on 25 Sep 2008, 00:49, edited 1 time in total.
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12 Oct 2006, 05:07
Hi all

Here is something useful for these kind of questions.

Permutation with repetition

--------------------------------------------------------------------------------

When order matters and an object can be chosen more than once then the number of permutations is

n^r

where n is the number of objects from which you can choose and r is the number to be chosen.

For example, if you have the letters A, B, C, and D and you wish to discover the number of ways to arrange them in three letter patterns (trigrams) you find that there are 4^3 or 64 ways. This is because for the first slot you can choose any of the four values, for the second slot you can choose any of the four, and for the final slot you can choose any of the four letters. Multiplying them together gives the total.
Permutation with repetition   [#permalink] 12 Oct 2006, 05:07
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