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A certain stock exchange designates each stock with a 1, 2 [#permalink]
10 Nov 2009, 14:50

2

This post was BOOKMARKED

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A

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E

Difficulty:

35% (medium)

Question Stats:

63% (02:18) correct
37% (01:47) wrong based on 180 sessions

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?

Re: Permutation Problem [#permalink]
10 Nov 2009, 14:57

1

This post received KUDOS

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

Re: Permutation Problem [#permalink]
10 Nov 2009, 14:59

6

This post received KUDOS

Expert's post

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 =2 6^2 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. _________________

Re: Permutation Problem [#permalink]
31 Jan 2011, 08:48

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

Re: Permutation Problem [#permalink]
14 Feb 2013, 08:18

1

This post received KUDOS

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.

Re: Permutation Problem [#permalink]
15 Feb 2013, 02:51

2

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

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, ...

Re: Permutation Problem [#permalink]
18 Sep 2013, 02:56

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" _________________

Re: Permutation Problem [#permalink]
18 Sep 2013, 03:08

1

This post received KUDOS

Expert's post

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.

Re: Permutation Problem [#permalink]
18 Jan 2014, 05:48

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.

gmatclubot

Re: Permutation Problem
[#permalink]
18 Jan 2014, 05:48