Application of conditional combinatorics 
How many license plates can be made using $3$ letters of the English alphabet and $3$ digit number where:
  
  
*
  
*The first digit is not zero.
  
*Repetition of letters and digit is not allowed.
  

Is:
$$
\frac{n!}{(n-r)!}
$$
the right formula? I got $720$ well I'm not sure of it, I just solved $6$! 
 A: We will assume that the plate has shape three letters followed by three digits, as in MSE 901
The first letter can be chosen in $26$ ways. For each of these choices, the second letter can be chosen in $25$ ways. And for every one of those ways, the third letter can be chosen in $24$ ways. So the "letter" part of the plate can be chosen in $(26)(25)(24)$ ways.
The first digit of the digits part can be chosen in $9$ ways, since we must avoid $0$. For each of these ways, the second digit can be chosen in $9$ ways (we must avoid the first digit). And finally for each choice of first two digits, the third can be chosen in $8$ ways, for a total of $(9)(9)(8)$.
The total number of plates is therefore $(26)(25)(24)(9)(9)(8)$.
A: HINT: I'm assuming that you mean that the three letters precede the three digits, so that the plate has the form LLLNNN. There are $26$ choices for the first letter. Once it's been chosen, there are only $25$ possible choices for the second letter, since it is not allowed to repeat the first letter. Once the first two letters have been chosen, there are just $24$ possibilities for the third letter. Thus, by the multiplication principle (or Chinese menu principle) there are $26\cdot25\cdot24$ ways to choose the three letters. Now continue the analysis in the same way to complete the calculation. How many ways are there to choose the first digit? Once it's been chosen, how many ways are the to choose the second digit? And once those have been chosen, how many ways are there to choose the last digit?
