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April 16th, 2013, 02:15 PM   #1
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weird probability problem...

Suppose Texas l;icense plates consist of 7 characters. The first two being letters of the alphabet. The next 3 being numbers and the last two being letters. Example of a license plate: AB123CD
Determine the total number that can be created if.....
a) There are no restrictions
b) No letter may be repeated
c) The digits in characters 4 and 5 may not contain the digit zero
d) the digits in characters 4 and 5 may no contain the digit zero and the letter 0 may not be used.

This whole problem is a little intimidating for me. This is part of the review for my test tomorrow. I had to pic my sister up from the air port so my buddy sent me a pic of the review, but he wasn't paying attention so he didn't get any of the answers.... anyways. Here's me taking a stab at this.
a) c(26,2) c(10, 3) c(26, 2)
b) c(26,2) c(10,3) c(24,2)
c) c(26,2) c(10, 1) c(9,2) c(26, 2)
d) c(25,2) c(10,1) c(9,2) c (25,2)
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April 16th, 2013, 02:54 PM   #2
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Re: weird probability problem...

You are making a fundamental error (I assume that c( , ) means combination). You should not be dividing by 2, since order matters. For example: the answer to (a) is 26^2 x 10^3 x 26^2.
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April 16th, 2013, 02:58 PM   #3
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Re: weird probability problem...

Hmmm thanks for the insight mathman!

So what you're saying is
b) 25^2 X 10^3 X 24^2
c) 26^2 x 10^1 x 9^2 x 26^2
d) 25^2 x 10^1 x 9^2 x 25^2
?????


And yes I was using the combination formula. Probably would have been wise for me to mention that..
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April 16th, 2013, 03:15 PM   #4
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Re: weird probability problem..

I would simply use the fundamental counting principle to answer these questions.

a) How many character choices do you have for each digit?
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April 16th, 2013, 03:20 PM   #5
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Re: weird probability problem...

I have merged your duplicate topics...you see what happened because you posted twice? I wasted my time, and my time is valuable, at least to me.

Please pick the most appropriate sub-forum for your topic and post it only once. This avoids redundancy and duplication of effort on the part of our contributors.
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April 16th, 2013, 04:10 PM   #6
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Re: weird probability problem...

Sorry about that mark :/

I posted it in high school section first and figured that not many high schoolers would respond.

Won't happen again, promise
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April 17th, 2013, 12:03 PM   #7
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Re: weird probability problem...

Quote:
Originally Posted by hamburgertime
Hmmm thanks for the insight mathman!

So what you're saying is
b) 25^2 X 10^3 X 24^2
c) 26^2 x 10^1 x 9^2 x 26^2
d) 25^2 x 10^1 x 9^2 x 25^2
?????


And yes I was using the combination formula. Probably would have been wise for me to mention that..
c) and d) look ok. b) should be 26 x 25 x 10^3 x 24 x 23.
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