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-   -   License Plate (http://www.physicsforums.com/showthread.php?t=452933)

kingtaf Dec1-10 09:49 PM

License Plate
 
A witness to a hit-and-run accident tells the police that the license plate of the car
in the accident, which contains three letters followed by three digits, starts with the latters AS and contains both the digits 1 and 2. How many different license plates fit the description?

dmatador Dec2-10 12:09 AM

Re: License Plate
 
I shall start with the letters section of the plate:

Since you know the first two letters, the last letter comes from a set of 26 letters, so you have 26 choices, that is 1 x 1 x 26 different ways to have this first alphabetical section of the plate. Next comes the section with numbers. 1 and 2 must be among the three numbers, and you have 10 choices for the following space. The tricky part here is that 1 and 2 can be in any arrangement within the three spaces, and there are precisely 6 ways to have these two prescribed numbers in the three sections on the plate. The last spot will be chosen from a set of 10 number (0 - 9). This means that, in total, there are 1 x 1 x 26 x 1 x 1 x 6 different license plate arrangements with the given requirements.

Edit: I did this wrong. For the numbers section, you have 6 basic choices (let '(10)' denote the place among the three numbers on the plate that can take any of the ten values from 0 - 9): 12(10), (10)12, 1(10)2, 21(10), (10)21, 2(10)1. Each one of these basic choices consists of ten unique arrangements of the three numbers, and each of these ten sets can be paired with the 26 different arrangements of the letters section of the plate. So the total number of plates is 26 x 10 x 6. Is this right? I'm tired.

awkward Dec2-10 08:34 PM

Re: License Plate
 
The letters portion is easy; there are 26 choices.

The sequence of three digits containing at least one 1 and at least one 2 is a little trickier. One way to approach it is to break the possibilities into three cases:
1. Arrangements of 1, 2, x where x is a digit other than 1 or 2
2. Arrangements of 1, 1, 2
3. Arrangements of 1, 2, 2
Add up the number of possibilities in the three cases, multiply by the 26 possibilities for the letters, and that should do it.


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