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Q5: If a code word is defined to be a sequence of different

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Manager
Joined: 11 Jan 2007
Posts: 202
Location: Bangkok
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Q5: If a code word is defined to be a sequence of different [#permalink]  05 Jun 2007, 19:26
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Q5:
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, and J, what is the ratio 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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cool

Director
Joined: 26 Feb 2006
Posts: 905
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Kudos [?]: 52 [0], given: 0

Re: PS [#permalink]  05 Jun 2007, 19:54
jet1445 wrote:
Q5: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, and J, what is the ratio 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

E. without repetation
= (10 x 9 x 8 x 7 x 6) /(10 x 9 x 8 x 7) = 6:1

with repetation = (10^5)/ (10^4) = 10:1
Manager
Joined: 17 Oct 2006
Posts: 52
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Kudos [?]: 1 [0], given: 0

Its a permutation problem because the order does matter i.e., abcde is different than edcba. When we have decided that it is a permutation problem, then there is just simple math to do.
so from the data of the question
10P5 : 10P4
= 10!/(10-5)! : 10!/(10-4)!
= 10!/5! : 10!/6!
= 10*9*8*7*6*5!/5! :10*9*8*7*6!/6!
= 10*9*8*7*6 : 10*9*8*7
so 10*9*8*7*6/10*9*8*7
=6/1
=6:1 so E
Manager
Joined: 20 Dec 2004
Posts: 180
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It's a permutation problem
10P5/10P4 = 6:1
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Regards

Subhen

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