How many license plates can be made so that it begins with two letters and ends with five digits repetitions are allowed?

#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 \times 26 \times $ __ __ __ = ?

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 \times 25 \times $ __ __ __ = ?

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.

$26 \times 25 \times 10 \times 9$ __ = ?

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