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If a code word is defined to be a sequence of different
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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 5letter code words to the number of 4letter 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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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
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28 Jan 2012, 04:02
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 5letter code words to the number of 4letter 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
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23 Dec 2012, 13:29
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
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28 Dec 2012, 06:43
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 5letter code words to the number of 4letter code words?
A. 5 to 4 B. 3 to 2 C. 2 to 1 D. 5 to 1 E. 6 to 1 Number of 5letter code formed from 10 letters: \(=10*9*8*7*6\) Number of 4letter 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
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28 Dec 2012, 06:56
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 5letter code words to the number of 4letter 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 5letter code: 10!/5! = 10*9*8*7*6 Number of ways to form 4letter code: 10!/6! = 10*9*8*7 Ratio: 6 to 1 Answer : E
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Re: Ratio
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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 5letter code words to the number of 4letter 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 aresearcherplanstoidentifyeachparticipantinacertain134584.htmlCan you please when to use permutaions or Combinations in these type of problems?
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Re: Ratio
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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 5letter code words to the number of 4letter 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 aresearcherplanstoidentifyeachparticipantinacertain134584.htmlCan 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: aresearcherplanstoidentifyeachparticipantinacertain134584.html#p1150091
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Re: If a code word is defined to be a sequence of different
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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
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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: mathcombinatorics87345.html
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If a code word is defined to be a sequence of different
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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
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05 Jan 2018, 16:46
Hi All, To calculate the number of 5letter codes and 4letter 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 5letter codes = (10)(9)(8)(7)(6) Total 4letter codes = (10)(9)(8)(7) Notice how the number of 5letter codes is the total of 4letter codes multiplied by 6. Thus, the ratio of codes is 6:1 Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: If a code word is defined to be a sequence of different
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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 5letter code words to the number of 4letter 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
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15 May 2018, 09:22
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 5letter code words to the number of 4letter 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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Re: If a code word is defined to be a sequence of different
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05 Nov 2018, 09:18
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 5letter code words to the number of 4letter code words?
A. 5 to 4 B. 3 to 2 C. 2 to 1 D. 5 to 1 E. 6 to 1 10P5 /10P4 . Why we used Permutations not combinations as here sequence is important (Code)
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Re: If a code word is defined to be a sequence of different
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05 Nov 2018, 10:15
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 5letter code words to the number of 4letter code words?
A. 5 to 4 B. 3 to 2 C. 2 to 1 D. 5 to 1 E. 6 to 1 OA: E Number of \(5\) letter code words \(= 10*9*8*7*6\) Number of \(4\) letter code words \(= 10*9*8*7\) the ratio of the number of \(5\)letter code words to the number of \(4\)letter code words \(=\frac{ 10*9*8*7*6}{10*9*8*7}=\frac{6}{1}\)
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06 Nov 2018, 09:44
The ratio of the number of 5letter code words to the number of 4letter code words chosen from the 10 letters A, B, C, D, E, F, G, H, I, and J is given by 10P5/10P4 = 6:1 Option E is the correct Ans!!
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Re: If a code word is defined to be a sequence of different
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14 Apr 2019, 10:45
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 5letter code words to the number of 4letter code words?
A. 5 to 4 B. 3 to 2 C. 2 to 1 D. 5 to 1 E. 6 to 1 5 letter = 10*9*8*7*6 4 letter = 10*9*8*7 ratio ; 6 : 1 IMO E
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