A company assigns product codes consisting of all the letter

Question

A company assigns product codes consisting of all the letters in the alphabet.How many product codes are possible if the company uses at most 3 letters in its codes, and all letters can be repeated in any one code?

A.15600
B.16226
C.17576
D.18278
E.28572

A. 15600
B. 16226
C. 17576
D. 18278
E. 28572

BrentGMATPrepNow · Accepted Answer

1-letter codes  
26 letters, so there are 26 possible codes

2-letter codes  
There are 26 options for the 1st letter, and 26 options for the 2nd letter.
So, the number of 2-letter codes = (26)(26) = 26²

3-letter codes  
There are 26 options for the 1st letter, 26 options for the 2nd letter, and 26 options for the 3rd letter.
So, the number of 3-letter codes = (26)(26)(26) = 26³

So, the TOTAL number of codes = 26 + 26² + 26³

IMPORTANT  : Before we perform ANY calculations, we should first look at the answer choices, because we know that the GMAT test-makers are very reasonable, and they don't care whether we're able make long, tedious calculations. Instead, the test-makers will create the question (or answer choices) so that there's an alternative approach.

The alternative approach here is to recognize that:
26 has  6  as its units digit 
26² has  6  as its units digit 
26³ has  6  as its units digit

So, (26)+(26²)+(26³) = (2 6 )+(___ 6 )+(____ 6 ) = _____  8

Since only D has   8   as its units digit, the answer must be D

Cheers, 
Brent

mau5 · Answer

The no of ways in which the company can make codes of 1 letter : 2 6 
# for 2 letters : 26*26 = _ _  6 
# for 3 letters : _ _ 6 * 26 =_ _ _ _  6

The total : By adding all of them , the units digit would be : 6+6+6 = _ _ _ _ _  8

D.

Bunuel · Answer

Similar question from GMAT Prep: Discussed here: a-certain-stock-exchange-designates-each-stock-with-a-86656.html Other similar questions to practice: all-of-the-stocks-on-the-over-the-counter-market-are-126630.html if-a-code-word-is-defined-to-be-a-sequence-of-different-126652.html the-simplastic-language-has-only-2-unique-values-and-105845.html a-4-letter-code-word-consists-of-letters-a-b-and-c-if-the-59065.html a-certain-stock-exchange-designates-each-stock-with-a-86656.html a-5-digit-code-consists-of-one-number-digit-chosen-from-132263.html a-company-that-ships-boxes-to-a-total-of-12-distribution-95946.html a-company-plans-to-assign-identification-numbers-to-its-empl-69248.html the-security-gate-at-a-storage-facility-requires-a-five-109932.html all-of-the-bonds-on-a-certain-exchange-are-designated-by-a-150820.html a-local-bank-that-has-15-branches-uses-a-two-digit-code-to-98109.html a-researcher-plans-to-identify-each-participant-in-a-certain-134584.html baker-s-dozen-128782-20.html#p1057502 in-a-certain-appliance-store-each-model-of-television-is-136646.html m04q29-color-coding-70074.html john-has-12-clients-and-he-wants-to-use-color-coding-to-iden-107307.html how-many-4-digit-even-numbers-do-not-use-any-digit-more-than-101874.html a-certain-stock-exchange-designates-each-stock-with-a-85831.html Hope it helps.

bagdbmba · Answer

Thanks Bunuel for not only referring to the solution but also sharing several similar problems for practice.
Really appreciate it.

+1 Sir!

JeffTargetTestPrep · Answer

A one-letter code can be formed in 26 ways.
 
A two-letter code can be formed in 26 x 26 = 26^2 ways.
 
A three-letter code can be formed in 26 x 26 x 26 = 26^3 ways.
 
Thus, the total number of codes is:
 
26 + 26^2 + 26^3 
 
26(1 + 26 + 26^2)
 
26(1 + 26 + 676)
 
26(703) = 18,278
 
Notice that one need not do the multiplication out fully. We see that the units digit is an 8. 
 
Answer: D

shashankism · Answer

So product codes can be o f 1 letter or 2 letter or 3 letters and all the letters can be repeated in any code.
So product codes possible = 26 + 26*26 + 26*26*26 = 26(1+26+26^2) = 26*(27+676) = 26* (703) = 18278

Answer D

MT1302 · Answer

In 2 letter code and 3 letter code , the 2 codes or 3 codes can be similar for eg. AA or AAA , then why are we not dividing in case of AA or AAA by 2! & 3! respectively ?

Bunuel · Answer

With the method provided, you get one codes 'AA' and and one code 'AAA'. Why consider a factorial correction? Factorial division is used to eliminate duplicates. What would you exclude here? Dividing these by a factorial isn't necessary, as they are unique combinations, not duplicates.

MT1302 · Answer

In 2 letter code and 3 letter code , the 2 codes or 3 codes can be similar for eg. AA or AAA , then why are we not dividing in case of AA or AAA by 2! & 3! respectively ?

With the method provided, you get one codes 'AA' and and one code 'AAA'. Why consider a factorial correction? Factorial division is used to eliminate duplicates. What would you exclude here? Dividing these by a factorial isn't necessary, as they are unique combinations, not duplicates.

Ok, so if the question stem mentions that "AA" is identical then would we be diving by 2! , Correct ? 
 Bunuel

Bunuel · Answer

Not sure I follow. What do you mean by "if the question stem mentions that "AA" is identical "?

MT1302 · Answer

So if the Question stem is mentioning that "same letters for codes are identical " then what changes would we be making in the solution ?

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A company assigns product codes consisting of all the letter

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