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Combination problem

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Combination problem

by fightthegmat » Sat Sep 05, 2009 12:52 am
Q]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

Pls help with how we get the solution
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by real2008 » Sat Sep 05, 2009 1:13 am
it is (10*9*8*7*6)/(10*9*8*7)=6:1 (E)
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by Nermal » Sat Sep 05, 2009 1:30 am
@real2008 Can you please explain how you got there.

I get the answer 6:5 which unfortunately is no option among the answer choices.

I tell you how I got there and hope someone can tell me where it went wrong.

This is my way:

There are 10 letters of which you have to chose 5 respectively 4 letters from in order to form the code.

Therefore
no. of 5-letter-codes: 10C5 = (10*9*8*7*6)/(1*2*3*4*5) = 252

no. of 4-letter-codes: 10C4 =
(10*9*8*7)/(1*2*3*4) = 210

Now you divide 252/210 = 6/5
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by real2008 » Sat Sep 05, 2009 3:13 am
Nermal wrote:@real2008 Can you please explain how you got there.

I get the answer 6:5 which unfortunately is no option among the answer choices.

I tell you how I got there and hope someone can tell me where it went wrong.

This is my way:

There are 10 letters of which you have to chose 5 respectively 4 letters from in order to form the code.

Therefore
no. of 5-letter-codes: 10C5 = (10*9*8*7*6)/(1*2*3*4*5) = 252

no. of 4-letter-codes: 10C4 =
(10*9*8*7)/(1*2*3*4) = 210

Now you divide 252/210 = 6/5
u need form a code of 5 different letters. u have 10 different letters. Naturally first letter of the code can be chosen in 10 ways. second letter by 9 ways (because after choosing first letter u have 9 letters), third one by 8 ways and fourth one by 7 ways, fifth one by 6 ways

Hence total no of ways 10*9*8*7*6

similarly to form a code of four different letters, no of possible ways is 10*9*8*7

hence ratio 6:1
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by Nermal » Sat Sep 05, 2009 4:50 am
I found out where I made the mistake.
I treated the question as a combination's instead of as a permutation's problem.
The formula to this is: n!/(n-k)!

So here is my answer now:

5-letter-codes: 10!/(10-5)!=10!/5!=10*9*8*7*6*5*4*3*2*1/(5*4*3*2*1)=30240

4-letter-codes: 10!/(10-4)!=10!/6!=10*9*8*7*6*5*4*3*2*1/(6*5*4*3*2*1)=5040

30240/5040=6/1
Hence 6:1
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