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# If a code word is defined to be a sequence of different letters chosen

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Manager
Joined: 06 May 2009
Posts: 50
If a code word is defined to be a sequence of different letters chosen  [#permalink]

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09 Jul 2009, 01:08
00:00

Difficulty:

55% (hard)

Question Stats:

71% (00:26) correct 29% (00:50) wrong based on 91 sessions

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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

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-a-code-word-is-defined-to-be-a-sequence-of-different-126652.html
Senior Manager
Joined: 25 Mar 2009
Posts: 288
Re: If a code word is defined to be a sequence of different letters chosen  [#permalink]

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09 Jul 2009, 08:19
2
1
I’m not too good at these, but I’ll give it a shot. The wording of the question seems like order should not matter, but I tried solving that way and it didn't work. In this case ORDER MATTERS. I am unable to articulate why. Someone please explain for all of our edification.

Method #1: Using the permutation formula
# of 5 letter code words:
nPr = n! / (n – r)! = 10! / (10 – 5)! = 10*9*8*7*6

# of 4 letter code words:
nPr = 10! / (10 – 4)! = 10*9*8*7

# of 5 words / # of 4 words = (10*9*8*7*6)/(10*9*8*7) = 6 = 6/1

Method #2: Using the “slots” method

# of 5 letter code words:
There are 5 empty slots below. For the first slot, you have an option of choosing 10 different letters. So fill that slot in with 10. For the second slot, you now have 9 letters to choose from. For the third slot, you have 8 letters. Keep going until you fill in all the slots and multiply all the numbers. That gives you the total # of possibilities.
__ __ __ __ __

10 * 9 * 8 * 7 * 6

# of 4 letter code words:
__ __ __ __

10 * 9 * 8 * 7

Ratio = (10*9*8*7*6)/(10*9*8*7) = 6 = 6/1

Math Expert
Joined: 02 Sep 2009
Posts: 52211
Re: If a code word is defined to be a sequence of different letters chosen  [#permalink]

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13 Mar 2015, 20:34
ankur55 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 $$P^4_{10}$$;

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

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-a-code-word-is-defined-to-be-a-sequence-of-different-126652.html
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Joined: 09 Sep 2013
Posts: 9419
Re: If a code word is defined to be a sequence of different letters chosen  [#permalink]

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21 Jul 2018, 20:29
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Re: If a code word is defined to be a sequence of different letters chosen &nbs [#permalink] 21 Jul 2018, 20:29
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