## Code in alphabetical order

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### Code in alphabetical order

by imawolf » Wed Sep 13, 2017 7:38 am
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?

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### Code in alphabetical order

by Brent@GMATPrepNow » Wed Sep 13, 2017 7:50 am
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
Take the task of creating a 3-letter code and break it into stages.

Stage 1: Select 3 letters from the alphabet
Since the order in which we select the letters does not matter, we can use combinations.
We can select 3 letters from 26 letters in 26C3 ways (2600 ways)
So, we can complete stage 1 in 2600 ways

Stage 2: Arrange the 3 letters in alphabetical order
IMPORTANT: Once we have selected 3 letters in stage 1, this next step can be accomplished in only 1 way.
For example, if we selected G, B and R in stage 1, then there's only one way to arrange these letters alphabetically (BGR)
So we can complete stage 2 in 1 way

By the Fundamental Counting Principle (FCP), we can complete the 2 stages (and thus create a 3-letter code) in (2600)(1) ways ([spoiler]= 2600 ways[/spoiler])

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

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Cheers,
Brent
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by [email protected] » Wed Sep 13, 2017 6:13 pm
Hi imawolf,

What is the source of this question? I ask because it does not include many of the 'details' that an Official question would likely include. For example, it refers to 'three digit' codes when it's clearly meant to infer 'three-letter' codes. We're meant to infer that we're dealing with the 26-letter English alphabet, but the prompt never explicitly states that and we're also meant to infer that duplicate letters are NOT allowed (which the prompt also does not state).

Assuming all of those details though, then Brent's solution is spot-on. That having been said, if this lack of attention-to-detail is typical in the questions that came from this source, then you might want to work with a different resource that provides more reputable study materials.

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by imawolf » Wed Sep 13, 2017 6:24 pm
Very clear! Thanks Brent!

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by Matt@VeritasPrep » Wed Sep 20, 2017 4:09 pm
Let me add a step here, since calculating 26 choose 3 isn't the most natural thing in the world:

26 choose 3 =

26! / (23! * 3!) =

(26 * 25 * 24) / (3 * 2 * 1) =

26/2 * 24/3 * 25/1 =

13 * 8 * 25 =

13 * 200 =

2600

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by mbawisdom » Fri Mar 02, 2018 7:37 am
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.

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