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Pretty simple actually
Number of 1 letter symbols = 26
Number of 2 letter symbols = 26*26
Number of 3 letter symbols = 26*26*26

The answer is sum of the 3. However if you are a real GMAT club member and confident about your approach you won't calculate any further.

The answer choices have their units digits unique. All 3 of the above end in 6 (6 to the power of anything ends in 6) so 6+6+6 = 18 i.e. ends with 8.

So the answer is E.
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Remember that Numbers can be repeated.

Number of 1 Digit Codes = 26
Number of 2 Digit Codes = 26 * 26 = 676
Number of 2 Digit Codes = 26 * 26 * 26 = 17576

Total = 26 + 676 + 17576 = 18278

In future do not put OA with the question. As people might solve it back wards to reach the solution.
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The important language in this one is "letters may be repeated" (i.e. selection with replacement) and "letters used in a different order constitute a different code" (i.e. order doesn't matter).

There are 26 one-digit codes.
There are 26*26 two-digit codes.
There are 26*26*26 three-digit codes. (Note: not 26*25*24, because letters can repeat)

There several ways to do the actual calculation, but I did it this way:
26 + 26*26 + 26*26*26
26 (1 + 26 + 26^2)
26 (27 + 676)
26 (703)
18278

Answer (D) 16302 seems to correspond to the following: 26 + 26*26 + 26*25*24.
Answer (C) 15600 is 26+ 26^2 less than (D), so it corresponds to: 26*25*24.
Answer (B) 8125 is 25*25*13, which is suppose could be arrived at by doing 25*25*26/2.
Answer (A) 2951 is 13*227, and I'm not sure what set-up error would lead one to arrive at this.

The wrong answers are interesting to examine, because they reveal what errors the GMAT writers suspect people will make.
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esledge

In this question, don't you think that order is important. AB is a different code than BA. Therefore, shouldn't we use permutations instead of combinations?

I was expecting something like 26C1.26C1.2! for the two digit codes & 26c1.26C1.26C1.3! for 3 digit codes!

Please correct me!
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honeyrai
In this question, don't you think that order is important. AB is a different code than BA. Therefore, shouldn't we use permutations instead of combinations?

I was expecting something like 26C1.26C1.2! for the two digit codes & 26c1.26C1.26C1.3! for 3 digit codes!

Please correct me!
Let's look at the two digit codes:
A and B are both among the 26 letters from which you select the first digit: 26C1.
A and B are both among the 26 letters from which you select the second digit: 26C1.

Thus, with (26C1)(26C1) you are already including AB and BA (and AA and BB, etc.), so no need to increase the count with some multiplier for "shuffling."
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Bunuel
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A certain stock exchange designates each stock with a 1, 2, or 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 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) 18,278

1 letter code=26
2 letter code=26^2
3 letter code=26^3

Total=26+26^2+26^3

The problem we are faced now is how to get the answer quickly. Note that the units digit of 26+26^2+26^3 would be (6+6+6=18) 8. Only one answer choice has 8 as unit digit: E (18,278). So I believe, even not calculating 26+26^2+26^3, that answer is E.

Hi Bunuel,

Firstly let me say that i fully understand your explanation and it makes perfect sense. I am however, finding it difficult to understand why we can't plug in the numbers into the permutations formula i.e. 26+Pm26,2 + Pm26,3 =16,276 which is well short of the 18,278 answer. I'm just wondering when to apply the approach you mentioned above and when to apply the Permutations formula.

Thanks!
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Bunuel
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A certain stock exchange designates each stock with a 1, 2, or 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 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) 18,278

1 letter code=26
2 letter code=26^2
3 letter code=26^3

Total=26+26^2+26^3

The problem we are faced now is how to get the answer quickly. Note that the units digit of 26+26^2+26^3 would be (6+6+6=18) 8. Only one answer choice has 8 as unit digit: E (18,278). So I believe, even not calculating 26+26^2+26^3, that answer is E.

Hi Bunuel,

Firstly let me say that i fully understand your explanation and it makes perfect sense. I am however, finding it difficult to understand why we can't plug in the numbers into the permutations formula i.e. 26+Pm26,2 + Pm26,3 =16,276 which is well short of the 18,278 answer. I'm just wondering when to apply the approach you mentioned above and when to apply the Permutations formula.

Thanks!

Good question. +1.

Notice that we are told that the letters may be repeated, so AA, BBB, ACC, CAA, .... codes are possible.

Now, 26P2 is the number of ways we can choose 2 distinct letters out of 26 when the order matters, thus it doesn't account for the cases like AA, AAA, ABB, ...

Hope it's clear.
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japped187
A certain stock exchange designates each stock with a one-, two-, or three- 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 constitute 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

Answer is E, please explain
As discussed above, the solution is indeed 26 + 26*26 + 26*26*26. I noticed that all the digits were different in the options.
So I simply calculated the unit digit of each product. 6 +6 + 6. = 18 = Unit digit of answer = 8. Hence, E.
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lagomez
chicagocubsrule
A certain stock exchange designates each stock with a 1, 2, or 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 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) 18,278

Answer e.

if each letter is the same: 26 different combinations
2 letters the same 26^2
all different 26^3

26^3 + 26^2 + 26 = 18278

what does this statement exactly mean-
"if the same letters used in a different order constitute a different code"
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honchos
lagomez
chicagocubsrule
A certain stock exchange designates each stock with a 1, 2, or 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 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) 18,278

Answer e.

if each letter is the same: 26 different combinations
2 letters the same 26^2
all different 26^3

26^3 + 26^2 + 26 = 18278

what does this statement exactly mean-
"if the same letters used in a different order constitute a different code"

It means that the order of the letters matters. For example, code AB is different from BA.
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I dont know why , but I was thinking for one letter, it's 26,
Then for 2 same ones it would be 26^2
2 different ones would mean 26*25 * 2 (because a different order)
3 same would be 26^3, and 3 different would be 26*25*24*3!....Where am I (obviously) double counting?
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usre123
I dont know why , but I was thinking for one letter, it's 26,
Then for 2 same ones it would be 26^2
2 different ones would mean 26*25 * 2 (because a different order)
3 same would be 26^3, and 3 different would be 26*25*24*3!....Where am I (obviously) double counting?

How is 26^2 the number of two same letter words? How is 26^3 the number of three same letter words? Isn't both 26? AA, BB, CC, ..., ZZ and AAA, BBB, CCC, DDD, ..., ZZZ?

26^2 gives the number of ALL 2-letter words possible, the same way as 26^3 gives the number of ALL 3-letter words possible.
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Quote:
A certain stock exchange designates each stock with a one-, two- or three-letter code, where each letter is selected from the 26 letters of the alphabets. If the letter maybe repeated and if the same letters used in different order constitude a different code, how many different stock is it possible to uniquely designate with these codes?

A. 2,951
B. 8,125
C. 15,600
D. 16,302
E. 18,278

My approach is similar to that of Bhoopendra, with a TWIST at the end.

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) = (26)+(___6)+(____6) = _____8

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

Cheers,
Brent
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lagomez
chicagocubsrule
A certain stock exchange designates each stock with a 1, 2, or 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 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) 18,278

A 1-digit code can be created in 26 ways, a 2-digit code in 26^2 ways, and a 3-digit code in 26^3 ways.

Thus, the number of ways to create the 3 codes is:

26 + 26^2 + 26^3

We should recognize that 26, 26^2, and 26^3 all have units digits of 6. Thus, the sum of those 3 numbers will have a units digit of 8. The only answer choice that has a units digit of 8 is choice E. Thus, the answer must be 18,278.

Answer: E
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Bunuel, can you please explain the theory behind this problem. Why 26^2 and 26^3?
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ShivamoggaGaganhs
Bunuel, can you please explain the theory behind this problem. Why 26^2 and 26^3?

For a two-letter code the first letter can be any from 26, similarly the second letter can also be any from 26, thus the total number of two-letter codes is 26*26 = 26^2. You can apply the same logic for three-letter codes.
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Bunuel
A certain stock exchange designates each stock with a one-, two-, or three- 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 constitute 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

1 letter codes = 26
2 letter codes = 26^2
3 letter codes = 26^3

Total = 26 + 26^2 + 26^3

The problem we are faced now is how to get the answer quickly. Note that the units digit of 26+26^2+26^3 would be (6+6+6=18) 8. Only one answer choice has 8 as unit digit: E (18,278). So I believe, even not calculating 26+26^2+26^3, that answer is E.
Hi, should'nt we subtract 26 from 26^2 and 26^3 because AA,BB, CC or BBB,CCC,DDD will be occuring twice in the case?
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