If a code word is defined to be a sequence of different : GMAT Problem Solving (PS)
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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]

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05 Sep 2010, 05:52
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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
[Reveal] Spoiler: OA
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05 Sep 2010, 06:00
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Orange08 wrote:
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

# of ways to choose 5 different letters out of given 10 letters when the order of chosen letters matters (as for example code word ABCDE is different from BCDEA) is $$P^5_{10}$$;

# of ways to choose 4 different letters out of given 10 letters when the order of chosen letters matters is $$P^4_{10}$$;

$$Ratio=\frac{P^5_{10}}{P^4_{10}}=\frac{10!}{5!}*\frac{6!}{10!}=\frac{6}{1}$$.

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Re: If a code word is defined to be a sequence of different [#permalink]

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28 Dec 2012, 05:56
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Orange08 wrote:
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

Number of ways to form 5-letter code: 10!/5! = 10*9*8*7*6
Number of ways to form 4-letter code: 10!/6! = 10*9*8*7

Ratio: 6 to 1

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28 Nov 2012, 10:34
Just a question Bunnel, why should it be resolved as a permutation not combination ?
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28 Nov 2012, 22:45
hi Emaco,
If we solve it using combination, it would be something like this...

Select 5 alphabets from 10..so 10C5 and then arrange selected 5 alphabets in 5! ways. Therefore 10C5 * 5! ....(1)
Select 4 alphabets from 10..so 10C4 and then arrange selected 4 alphabets in 4! ways. Therefore 10C4 * 4! ....(2)

Take the ratio 1 / 2.

ⁿPr = (r!) · ⁿCr
However, are you aware of the formula?
nPr = nCr * r!

So either way you get the answer. Hope this helps.

Regards,
Pritish
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29 Nov 2012, 00:41
Thanks a lot for clarification, deeply appreciated !
Re: Code word sequence   [#permalink] 29 Nov 2012, 00:41
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