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

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23 Aug 2006, 19:36
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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
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23 Aug 2006, 20:01
different sequence of numbers imply that ABCDE is different from EDCBA

so using permutations:

10P5/10P4 = (10!/5! )/ (10!/6!) = 6 to 1

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23 Aug 2006, 20:14
gmatornot 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

10c5 = Number of ways of selecting 5 letter code words
= 10*9*8*7*6/5!

10c4 = Number of ways of selecting 4 letter code words
= 10*9*8*7*/4!

@ This point i Divided 10c5/10c4 to get 6/5 which is not one of the answer choices.

So went back and saw that after selecting the 5 digit or 4 digit letter code we could arrange them differently

so Number of ways of selecting 5 letter words and arrangiing them=
10c5 * 5! = 10*9*8*7*6

so Number of ways of selecting 4 letter words and arrangiing them=
10c4 * 4! = 10*9*8*7

So required ratio = 6/1

What is the source of this ?

E

Heman
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23 Aug 2006, 20:17
trivikram wrote:
(10C5*5!)/(10C4*4!) = 3/2

B it is

I believe this calculation comes out to 6/1

Heman
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24 Aug 2006, 15:22
heman wrote:
trivikram wrote:
(10C5*5!)/(10C4*4!) = 3/2

B it is

I believe this calculation comes out to 6/1

Heman

I do this types of mistakes all the time and screw myself up ....
24 Aug 2006, 15:22
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