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A certain stock exchange designates each stock with a 1,2 or

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A certain stock exchange designates each stock with a 1,2 or [#permalink] New post 19 Dec 2007, 05:09
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 repeatyed and if the same letters used in a different order constitute a different code, how many different stocks is it possibile to uniquely designate with these codes?

A.2951
B.8125
C.15600
D. 16302
E.18278
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 [#permalink] New post 19 Dec 2007, 06:32
Expert's post
E

N=26+26^2+26^3=18278.

The shortcut for calculation:

last digit of 26: 6
last digit of 26^2: 6
last digit of 26^3: 6
last digit of the sum: 8

Therefore, E
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 [#permalink] New post 19 Dec 2007, 06:33
26+26*26+26*26*26=18278

Answer E
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 [#permalink] New post 19 Dec 2007, 07:05
I thought: 1c26+2c26+3c26...where's my mistake?
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 [#permalink] New post 19 Dec 2007, 07:10
Expert's post
marcodonzelli wrote:
I thought: 1c26+2c26+3c26...where's my mistake?


Your way has two mistakes:

1. " the letters may be repeatyed" leads to 26*26*26..... instead of 26*25*24 for nPm-nCm approach

2. "the same letters used in a different order constitute a different code" would lead rather nPm than nCm.
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Re: permutations - DIFFICULT [#permalink] New post 19 Dec 2007, 13:20
marcodonzelli 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 repeatyed and if the same letters used in a different order constitute a different code, how many different stocks is it possibile to uniquely designate with these codes?

A.2951
B.8125
C.15600
D. 16302
E.18278



This wasn't too difficult, just may seem so b/c we don't usually encounter these types of perm problems..

26 for 1 letter codes

26*26 for 2 letter codes

26*26*26 for three letter codes

26+676+17576=18278

E
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Re: permutations - DIFFICULT [#permalink] New post 21 Dec 2007, 21:19
marcodonzelli 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 repeatyed and if the same letters used in a different order constitute a different code, how many different stocks is it possibile to uniquely designate with these codes?

A.2951
B.8125
C.15600
D. 16302
E.18278


If repetition is allowed:

26 + 26*26 + 26*26*26

I am not sure how to calculate it faster.
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 [#permalink] New post 21 Dec 2007, 21:21
walker wrote:
E

N=26+26^2+26^3=18278.

The shortcut for calculation:

last digit of 26: 6
last digit of 26^2: 6
last digit of 26^3: 6
last digit of the sum: 8

Therefore, E


This short cut for calculation is awesome! Thanks Walker!
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 [#permalink] New post 22 Dec 2007, 00:56
LM wrote:
walker wrote:
E

N=26+26^2+26^3=18278.

The shortcut for calculation:

last digit of 26: 6
last digit of 26^2: 6
last digit of 26^3: 6
last digit of the sum: 8

Therefore, E


This short cut for calculation is awesome! Thanks Walker!


Walker,
There is a gap in my knowledge of this problem. I can understand only the case 26*26*26. B/c the codes can be designate by 26C1*26C1*26C1. Can Walker make clear why I must plus 26 and 26*26?

Many thanks
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 [#permalink] New post 22 Dec 2007, 01:39
Expert's post
sondenso wrote:
Walker make clear why I must plus 26 and 26*26?


"A certain stock exchange designates each stock with a 1,2 or 3-letter code"

You should calculate separately for 1-letter code, 2-letter code and 3-letter code and sum them.
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 [#permalink] New post 22 Dec 2007, 01:53
The solution will be 26+26*26+26*26*26
The shortcut can look like this 26+26*26+26*26*26=26*(1+26+26*26)=
=26*(1+26*(1+26))=26*(1+26*27)=18278

The advantage is that it is not a guess-like strategy, but a more reliable one.

Answer: E[/quote]
  [#permalink] 22 Dec 2007, 01:53
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