GMAT Math

imawolf wrote:A store labels its products with three digits codes composed entirely of letters of the alphabet arranged alphabetically. How many such codes can be created?

(A) 2600
(B) 2936
(C) 9630
(D) 14526
(E) 17576

Totally puzzled. How to solve this?

PRESUMPTION: you cannot use more than one letter twice

Letters in alphabet = 26

First, lets keep it simple, how many ways can we order 3 letters: If we use the slot method ____ ____ ____, in the first slot we can put 26 letters, second slot 25 letters and third slot 24 letters --> so 26*25*24

Constraint: must be alphabetical, okay so if this is the case only 1 out of 6 permutations are valid:
Lets take a look at the following possibilities when we get an A, B and C (only ABC is in alphabetical order):
ABC
ACB
BAC
BCA
CAB
CBA

Thus, we need to divide 26*25*24 by 6 to get the possible codes

Codes = 26*25*24 /6
Codes = 26*25*4
Codes = 26*100
Codes = 2600

WHAT IF THE PRESUMPTION ISN'T TRUE??? Then we could have code such as AAA, BBB, ABB, AAB etc

Codes with 3 letters all the same (AAA, BBB etc): 26
Codes with one letter and then two the same (ABB, ACC, ADD etc): if A at the beginning 25 combos (since you cant have A at beginning), B at the beginning 24 combos (since you cant have A/B at beginning), ... you should get the trend here... we need to sum up 1 to 25, (25*26/2), 325.
Codes with two letters the same and then one letter (AAB, BBC, BBD, CCD etc): same logic as above will give us 325.

If the presumption isn't true then we need to add another 676 codes, but that isn't a possible answer so we should assume that the presumption holds.