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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, 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?
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 \(P^4_{10}\);
Re: If a code word is defined to be a sequence of different
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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 -> \(10 * 9 * 8 * 7\)
Re: If a code word is defined to be a sequence of different
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28 Dec 2012, 05:43
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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: \(=10*9*8*7\)
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28 Dec 2012, 05:56
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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
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 \(P^4_{10}\);
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 \(P^4_{10}\);
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).
If a code word is defined to be a sequence of different
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18 Mar 2016, 01:42
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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, 15:46
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
Re: If a code word is defined to be a sequence of different
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15 May 2018, 07:48
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?
Re: If a code word is defined to be a sequence of different
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15 May 2018, 08:22
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?
_________________
Re: If a code word is defined to be a sequence of different
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05 Nov 2018, 08:18
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 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
10P5 /10P4 .
Why we used Permutations not combinations as here sequence is important (Code)
Re: If a code word is defined to be a sequence of different
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05 Nov 2018, 09:15
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 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
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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If a code word is defined to be a sequence of different
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06 Nov 2018, 08:44
The ratio of the number of 5-letter code words to the number of 4-letter 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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If a code word is defined to be a sequence of different &nbs
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06 Nov 2018, 08:44
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