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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28 Jan 2012, 02:59
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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

Last edited by Bunuel on 28 Jan 2012, 03:02, edited 1 time in total.
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28 Jan 2012, 03:02
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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

Notice that as we are dealing with code words then the order of the letters matters.

# 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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23 Dec 2012, 12:29
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Numbers of Options applicable for 5 letter digit -> $$10 * 9 * 8 * 7 * 6$$
( as option pool for first digit is 10, for second 9 because one is removed and so on)
Numbers of Options applicable for 5 letter digit -> $$10 * 9 * 8 * 7$$

Required Ratio -> $$(10 * 9 *8 * 7 * 6)/(10 * 9* 8 * 7)$$ = $$6:1$$
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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, 05:43
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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

Number of 5-letter code formed from 10 letters: $$=10*9*8*7*6$$
Number of 4-letter code formed from 10 letters: $$=10*9*8*7$$

Answer: 6 to 1 or E
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30 Mar 2013, 08:04
Bunuel 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

Notice that as we are dealing with code words then the order of the letters matters.

# 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 $$P^4_{10}$$;

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

Hi Bunnel,

In this problem you have used Permutations, but in the problem you have used combination, which also deals with code

a-researcher-plans-to-identify-each-participant-in-a-certain-134584.html

Can you please when to use permutaions or Combinations in these type of problems?
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31 Mar 2013, 07:51
mydreammba wrote:
Bunuel 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

Notice that as we are dealing with code words then the order of the letters matters.

# 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 $$P^4_{10}$$;

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

Hi Bunnel,

In this problem you have used Permutations, but in the problem you have used combination, which also deals with code

a-researcher-plans-to-identify-each-participant-in-a-certain-134584.html

Can you please when to use permutaions or Combinations in these type of problems?

In this case the order of the letters matters, but in other question we are only interested in codes which are in alphabetical order (so we are interested in only one particular order).

This post might help: a-researcher-plans-to-identify-each-participant-in-a-certain-134584.html#p1150091
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Re: If a code word is defined to be a sequence of different [#permalink]

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16 Apr 2014, 21:24
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Re: If a code word is defined to be a sequence of different [#permalink]

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23 Aug 2014, 09:51
Bunuel 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

Notice that as we are dealing with code words then the order of the letters matters.

# 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 $$P^4_{10}$$;

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

Hi Bunuel,

A couple of clarifications as i'm going through similar problems:

1) In this case, we use permutations because order DOES NOT matter. Correct?
2) If the problem said that we had to use the letter in alphabetical order, then order would matter and we would have to use the Combination formula over permutation. Correct?
3) Permutation and Combination both assumes that the letters/numbers CANNOT be repeated. Correct?
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Re: If a code word is defined to be a sequence of different [#permalink]

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31 Jan 2015, 06:46
For the life of me,

My calculation does not give me the same answer.

The different part is the 10 Choose 4, my calculations are giving me different figures??

When i factorize top and bottom, i get 252/ 210??

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

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31 Jan 2015, 07:10
Tmoni26 wrote:
For the life of me,

My calculation does not give me the same answer.

The different part is the 10 Choose 4, my calculations are giving me different figures??

When i factorize top and bottom, i get 252/ 210??

Thanks

It's not clear what you mean... Anyway, 10C4 = 10!/(6!4!) = 210
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31 Jan 2015, 07:18
Hello Bunuel,

Thank you very much for responding.

My problem with your initial calculation is that when i do understand how you got 6! in the bottom part of your example above.

When i calculate it, it goes 10 Choose 5 is = 10*9*8*7*6 / 5*4*3*2*1

And then 10 Choose 4 i= 10*9*8*7 / 4*3*2*1

when I calculate 10 Choose 5 / 10 Choose 4.... I get 6/5

I do not know where the error is coming from

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

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31 Jan 2015, 07:23
Tmoni26 wrote:
Hello Bunuel,

Thank you very much for responding.

My problem with your initial calculation is that when i do understand how you got 6! in the bottom part of your example above.

When i calculate it, it goes 10 Choose 5 is = 10*9*8*7*6 / 5*4*3*2*1

And then 10 Choose 4 i= 10*9*8*7 / 4*3*2*1

when I calculate 10 Choose 5 / 10 Choose 4.... I get 6/5

I do not know where the error is coming from

Thanks alot

$$P^5_{10}=\frac{10!}{5!}$$

$$P^4_{10}=\frac{10!}{6!}$$

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

Theory on Combinations: math-combinatorics-87345.html
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Re: If a code word is defined to be a sequence of different [#permalink]

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31 Jan 2015, 07:30
Once again, thanks for your response,

My main issue is with the 10 Choose 4 (Denominator) , my understanding is that in order to perform this calculation, I take the first 4 terms of 10! starting with 10 and divide that by 4 factorial.

I do not understand how you get 6! here

Thanks once again and i am very grateful for your time here
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31 Jan 2015, 07:32
Tmoni26 wrote:
Once again, thanks for your response,

My main issue is with the 10 Choose 4 (Denominator) , my understanding is that in order to perform this calculation, I take the first 4 terms of 10! starting with 10 and divide that by 4 factorial.

I do not understand how you get 6! here

Thanks once again and i am very grateful for your time here

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31 Jan 2015, 07:33
I am so so sorry, I was doing a Combination instead of a Permutation

Thank you very much
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07 Feb 2015, 10:31
In this case ABCDE is different from BCDEA hence order matters boils down to using permutations ...

Hence ( 10! / 5 ! ) * ( 6! / 10 ! ) = 6/1 Answer is E
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If a code word is defined to be a sequence of different [#permalink]

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18 Mar 2016, 01:42
Take 1 case: ABCDE for 5 letter code and ABCD for 4 letter code. You Choose 5 at a time and 4 at a time and order is important. So the formula is nPr. Therefore the ratio is 10P5/ 10P4 = 6:1
Hence E.
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Re: If a code word is defined to be a sequence of different [#permalink]

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04 Nov 2016, 01:52
Hi, i am confused why we dont care about the order of selection of 5 characters first then 4 characters or 4 characters first then 5 characters. My work is as below, can someone point out where did i do wrong?

First selection:
Ways of 5 characters to be selected in the beginning: 10 x 9 x 8 x 7 x 6 = 30240
Ways of 4 characters to be selected then: 5 x 4 x 3 x 2 =120

Second selection:
Ways of 4 characters to be selected first: 10 x 9 x 8 x 7 = 6860
Ways of 5 characters to be selected then: 6 x 5 x 4 x 3 x 2 = 720

So total of ways for 5 characters: 30960
total of ways for 4 characters: 6980

Ratio 5 characters: 4 characters = 4.436??
Re: If a code word is defined to be a sequence of different   [#permalink] 04 Nov 2016, 01:52
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