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How many vehicle license plates can be made if the licenses contains 2 letters of the English alphabet followed by a three digit number.

  1. If repetitions are allowed.
  2. If repetitions are not allowed.
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    i have a suspicion about the 1st question whether it will be 26*26*999 Or 26*26*1000.........I am not getting why the text book is used 9992011-04-22

2 Answers 2

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#1

The licence plate is described in-order so you can count the possibilities in order. Here is your plate:

__ __ __ __ __

What can go in the first spot? (ABCDEFGHIJKLMNOPQRSTUVWXYZ) So, there are 26 possibilities. What about the next spot? Again (ABCDEFGHIJKLMNOPQRSTUVWXYZ) .. so 26 again:

26 * 26 * __ __ __ = ?

now use the same idea for the spaces with numbers.


#2

Next, how do we count plates when you cannot repeat? If we have picked one of the 26 letters of the alphabet for the first element... then how many choices do we have for the next?

26 * 25 * __ __ __ = ?

Now, for the numbers. We have 10 possibilities as in the previous problem. (0123456789) But after we have used one number for the next slot there will only be 9 possibilities.

26 * 25 * 10 * 9 __ = ?

In fact, what we are doing is using a formula known as the "choose formula" or the binomial coefficient.

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    Although to be fair, O and Q can be hard to distinguish from 0, and I is hard to distinguish from 1, so our ex$p$ected number of ways to letter a license plate would be lower. A good exercise for the poster is to find how many ways to letter a license plate with these restrictions in place.2011-04-22
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1) If repetitions are allowed, then you have $26^2$ options to pick 2 letters, and $10^3$ to pick three digits. Total of $26^2\cdot 10^3$

2) If repititions are not allowed, then $\binom{26}{2}$ options to pick 2 letters, and $\binom{10}{3}$ to pick three digits. Total of $2\cdot\binom{26}{2}\times3!\cdot \binom{10}{3}$

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    2)If No repetition then 26P2*10P3=4680002011-04-22