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AND vs. OR - Combinations

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by Lifetron » Thu Apr 11, 2013 4:43 am
Each stack is designated with a 1, 2, and 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 constitutes a different code, how many different stocks is it possible to uniquely designate with these codes?
a. 2951
b. 8125
c. 15600
d. 16302
e. 18278

OA E

The answer will be valid only if the question is like this right - 1 or 2 or 3 letters. I was not able to get the question properly.

What am I missing ?

Thank you !
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by Anju@Gurome » Thu Apr 11, 2013 5:10 am
gughanbose wrote:The answer will be valid only if the question is like this right - 1 or 2 or 3 letters. I was not able to get the question properly.
You are correct.
In its current form the question can be interpreted as each stack is designated with a 1-letter, a 2-letter, and a 3-letter code.

However, the question actually meant to say '1 or 2 or 3 letter code'.
And I'll post the solution according to that as the solution involves a trick which can be used in some other problems.

Number of stacks with 1 letter code = 26
Number of stacks with 2 letter code = 26²
Number of stacks with 2 letter code = 26³
So, total number of different stacks = 26 + 26² + 26³

Now, how does one calculate such a large number?
The trick is you don't need to determine the exact value.
Units digit of 26 is 6. Hence, units digit of any power of 26 will be 6.
Hence, the units digit of the correct answer = units digit of (6 + 6 + 6) = 8

The correct answer is E.
Anju Agarwal
Quant Expert, Gurome

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by hooliajulia » Thu Apr 11, 2013 11:09 am
Hi,

This problem seems as if it would be a permutation problem. Why is this not the case?

Thank you,

Julia
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by th3509 » Thu Apr 11, 2013 11:14 am
Last edited by th3509 on Mon Apr 22, 2013 1:35 pm, edited 1 time in total.
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by Lifetron » Thu Apr 11, 2013 6:08 pm
hooliajulia wrote:Hi,

This problem seems as if it would be a permutation problem. Why is this not the case?

Thank you,

Julia
All possible combinations are covered in this -> 26 + 26² + 26³

It includes AB and BA... Ya, basically, it is a permutation problem.

Sorry, I mentioned it as combinations !
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