How many license plates can be made consisting of 2 letters followed by 3 digits (using the fundamental counting principle to solve)?

What I know: 26 letters in alphabet, so that means $2 \times 26$ 10 digits possible (0-9), so that means $3 \times 10$ FC principle says given $m$ and $n$ options gets you $m \times n$ varieties... ... However, the answer key says "676,000" when I got 1560...

Jul 11, 2017

$26 \times 26 \times 10 \times 10 \times 10 = 676 , 000$ possibilities

Explanation:

There is nothing stating that the letters and numbers can't be repeated, so all $26$ letters of the alphabet and all $10$ digits can be used again.

If the first is A, we have $26$ possibilities:
AA, AB, AC,AD,AE ...................................... AW, AX, AY, AZ.

If the first is B, we have $26$ possibilities:
BA, BB, BC, BD, BE .........................................BW, BX,BY,BZ

And so on for every letter of the alphabet.

There are $26$ choices for the first letter and $26$ choices for the second letter. The number of different combinations of $2$ letters is:
$26 \times 26 = 676$

The same applies for the three digits.
There are $10$ choices for the first, $10$ for the second and $10$ for the third:

$10 \times 10 \times 10 = 1000$

So for a license plate which has $2$ letters and $3$ digits, there are:

$26 \times 26 \times 10 \times 10 \times 10 = 676 , 000$ possibilities.

Hope this helps.