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A certain stock exchange designates each stock with a one-, two- or th

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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 diff stocks is it possible to designate with these codes?

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

[Reveal] Spoiler:
They say that QA is E, however I am not sure how to get there.

I have C 26 taken by 3 = 2,600
C of 26 taken by 2 = 325
C of 26 taken by 1 = 26.
If I add these, I get answer A. Not sure what I am doing wrong.

Thank you,
Andreea
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New post 26 Oct 2009, 13:57
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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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ralucaroman wrote:
HI,

Can smb please help me to understand the below problem?

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 diff stocks is it possible to designate with these codes?
A: 2,951
B: 8,125
C:15,600
D: 16,302
E: 18,278

They say that QA is E, however I am not sure how to get there.

I have C 26 taken by 3 = 2,600
C of 26 taken by 2 = 325
C of 26 taken by 1 = 26.
If I add these, I get answer A. Not sure what I am doing wrong.

Thank you,
Andreea


1 letter code - 26 (or you can write 26C1)
2 letter code - 26*26=26^2 (or you can write 26C1*26C1=26^2)
3 letter code - 26*26*26=26^3 (or you can write 26C1*26C1*26C1=26^3)

(For 2 letter, for example, you have 26 choices per each letter, it's 26^2 and not 26C2 or 26P2.)

Total =26+26^2+26^3 =18,278
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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

Last edited by chetan2u on 15 May 2016, 00:07, edited 3 times in total.
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chicagocubsrule wrote:
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

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chicagocubsrule wrote:
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 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.
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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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New post 08 Feb 2010, 04:00
Thanks.
I was doing 26+26.25+26.25.24
That would be correct if the letters could not be repeated, wouldnt be?

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Yes. Your answer will be correct in that way.

Also, I was just thinking since the last digit of the options are all different, we can just calculate the last digit for this instead of doing the actual multiplication and additions:
26 last digit: 6
26^2 last digit: 6
26^3 last digit: 6
Sum = 6+6+6 = x8 -- so answer should end in 8 -- option E

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New post 22 Feb 2010, 03:21
esledge wrote:
Great shortcut!

@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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New post 22 Feb 2010, 14:10
honeyrai wrote:
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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New post 01 Nov 2010, 19:19
The number of combinations for a stock w/ one letter is simply 26 (26 letters).

The number of combinations for a stock w/ two letters is 26*26 = 676.

The number of combinations for a stock w/ three letters is 26*26*26 = 17576.

Summing all of the possible combinations results in 17576 + 676 + 26 = 18278, hence answer E.

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Re: A certain stock exchange designates each stock with a one-, two- or th [#permalink]

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New post 31 Jan 2011, 09:48
Bunuel wrote:
chicagocubsrule wrote:
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.


The OA is E. Thanks for poininting out how to spot the correct answer - it took me miserable 4 minutes to multiply 26*26*26 and still I made a wrong calculation :oops:

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Bunuel wrote:
chicagocubsrule wrote:
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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iwillbeatthegmat wrote:
Bunuel wrote:
chicagocubsrule wrote:
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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New post 15 Feb 2013, 04:04
Perfectly clear! The repetition disqualifies the permutations formula.

Thanks alot Bunuel!

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Re: A certain stock exchange designates each stock with a one-, two- or th [#permalink]

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New post 18 Sep 2013, 03:56
lagomez wrote:
chicagocubsrule wrote:
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 wrote:
lagomez wrote:
chicagocubsrule wrote:
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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Bunuel wrote:
ralucaroman wrote:
HI,

Can smb please help me to understand the below problem?

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 diff stocks is it possible to designate with these codes?
A: 2,951
B: 8,125
C:15,600
D: 16,302
E: 18,278

They say that QA is E, however I am not sure how to get there.

I have C 26 taken by 3 = 2,600
C of 26 taken by 2 = 325
C of 26 taken by 1 = 26.
If I add these, I get answer A. Not sure what I am doing wrong.

Thank you,
Andreea


1 letter code - 26 (or you can write 26C1)
2 letter code - 26*26=26^2 (or you can write 26C1*26C1=26^2)
3 letter code - 26*26*26=26^3 (or you can write 26C1*26C1*26C1=26^3)

(For 2 letter, for example, you have 26 choices per each letter, it's 26^2 and not 26C2 or 26P2.)

Total =26+26^2+26^3 =18,278


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Re: A certain stock exchange designates each stock with a one-, two- or th [#permalink]

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New post 18 Jan 2014, 06:48
iwillbeatthegmat wrote:
Bunuel wrote:
chicagocubsrule wrote:
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!


1 letter code: 26
2-letter code: P(26,2) + 26 {P(26,2): 2 different numbers and different orders; 26: 2 same numbers}
3-letter code: P(26,3) + P(26, 2)C(3, 1) + 26 {P(26,3): 3 different numbers and different orders; P(26, 2)C(3, 1): 2 different numbers, one of which repeats; 26: 3 same numbers}

Hope it helps to understand.

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Re: A certain stock exchange designates each stock with a one-, two- or th   [#permalink] 18 Jan 2014, 06:48

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