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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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New post Updated on: 28 Jan 2012, 04:02
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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

Originally posted by RadhaKrishnan on 28 Jan 2012, 03:59.
Last edited by Bunuel on 28 Jan 2012, 04:02, edited 1 time in total.
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Re: Ratio  [#permalink]

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New post 28 Jan 2012, 04:02
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RadhaKrishnan 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 \(P^4_{10}\);

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

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

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New post 23 Dec 2012, 13:29
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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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New post 28 Dec 2012, 06:43
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RadhaKrishnan 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 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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Re: If a code word is defined to be a sequence of different  [#permalink]

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New post 28 Dec 2012, 06:56
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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

Answer : E
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Re: Ratio  [#permalink]

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New post 30 Mar 2013, 09:04
Bunuel wrote:
RadhaKrishnan 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 \(P^4_{10}\);

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

Answer: E.


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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Re: Ratio  [#permalink]

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New post 31 Mar 2013, 08:51
mydreammba wrote:
Bunuel wrote:
RadhaKrishnan 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 \(P^4_{10}\);

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

Answer: E.


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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New post 31 Jan 2015, 08:18
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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New post 31 Jan 2015, 08:23
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


Please use correct notations...

\(P^5_{10}=\frac{10!}{5!}\)

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

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

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

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New post 18 Mar 2016, 02:42
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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New post 05 Jan 2018, 16:46
Hi All,

To calculate the number of 5-letter codes and 4-letter codes, we have to set up 2 permutations. There's a 'shortcut' though - since the answer choices are RATIOS, we don't actually have to calculate the total number of each type of code.

Total 5-letter codes = (10)(9)(8)(7)(6)
Total 4-letter codes = (10)(9)(8)(7)

Notice how the number of 5-letter codes is the total of 4-letter codes multiplied by 6. Thus, the ratio of codes is 6:1

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

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New post 15 May 2018, 08:48
Bunuel wrote:
RadhaKrishnan 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 \(P^4_{10}\);

Answer: E.



Hi Bunuel,

I struggle with the the order of \(P^5_{10}\). From the theory I understood that the permutation is defined as: \(P^n_{k}\) n being the set of distinct objects and k being the number of objects chosen. Why is this upside down?

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

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New post 15 May 2018, 09:22
1
panache67 wrote:
Bunuel wrote:
RadhaKrishnan 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 \(P^4_{10}\);

Answer: E.



Hi Bunuel,

I struggle with the the order of \(P^5_{10}\). From the theory I understood that the permutation is defined as: \(P^n_{k}\) n being the set of distinct objects and k being the number of objects chosen. Why is this upside down?

Thanks!


\(P^5_{10}\), \(P^{10}_5\), 10P5 are all the same: choosing 5 out of 10, when the order matters. Just different ways of writing the same. Could it be choosing 10 out of 5?
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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