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If a code word is defined to be a sequence of different

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If a code word is defined to be a sequence of different [#permalink] New post 23 Jun 2006, 15:35
If a code word is defined to be a sequence of different letters chosen from the 10 letters A B C D E F G H I J what is the raatio of the number of 5 letter code words to the number of 4 letter code words.


A. 5 to 4
B 3 to 2
C 2 to 1
D 5 to 1
E 6 to 1
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 [#permalink] New post 23 Jun 2006, 16:04
As we are picking code-words this is permutations problem.

Given 10 alphabets, # of 5 letter codes = 10P5
# of 4 letter codes = 10P4

Ratio = (10!/5!)x (6!/10!) = 6:1

E. 6 to 1
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 [#permalink] New post 23 Jun 2006, 16:09
why is it multipled by 6!/10! makes no sense

wouldnt it be like

10C5 divided by 10 C 4

or 10*9*8*7*6/10*9*8*7

hence 6, oh i see...this would be easier to explain to people.
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 [#permalink] New post 23 Jun 2006, 16:10
how is this even right it says different letters, so doesnt that mean combination u cant pick the same letter twice.......

This problem is weird.
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 [#permalink] New post 23 Jun 2006, 16:27
Isn't this a Permutation problem not a combination?

nPr = n!/(n-r)!

10P4 = 10!/(10-4)! = 10!/6!

GMATCUBS21 wrote:
why is it multipled by 6!/10! makes no sense

wouldnt it be like

10C5 divided by 10 C 4

or 10*9*8*7*6/10*9*8*7

hence 6, oh i see...this would be easier to explain to people.
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 [#permalink] New post 23 Jun 2006, 18:34
10P5/10P6
E

courtesy haas_mba
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Last edited by gmatmba on 23 Jun 2006, 19:52, edited 1 time in total.
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 [#permalink] New post 23 Jun 2006, 18:43
Is that the equal to 10P5/10P4?

I think that comes to 6:5, not 6:1.

Did you mean 10P5/10P6?

gmatmba wrote:
10C5/10C6
E
  [#permalink] 23 Jun 2006, 18:43
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