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VP
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A certain stock exchange designates each stock with a 1,2 or [#permalink]
 19 Dec 2007, 06:09
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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CEO
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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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Senior Manager
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26+26*26+26*26*26=18278
Answer E
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VP
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I thought: 1c26+2c26+3c26...where's my mistake?
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CEO
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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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CEO
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Re: permutations - DIFFICULT [#permalink]
19 Dec 2007, 14:20
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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Director
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Re: permutations - DIFFICULT [#permalink]
21 Dec 2007, 22:19
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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Director
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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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SVP
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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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CEO
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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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Intern
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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]
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