A license plate has 6 characters. I found out that the number of license plates with 2 distinct letters and 4 distinct numbers is $\binom{26}{2}\binom{10}{4}6!$. How can I determine the number of license plates if the numbers are ordered and/or the letters are in alphabetical order ? I would like to count the following :
1) License plates with 2 distinct letters followed by 4 distinct numbers in increasing order
2) License plates with 2 distinct letters in alphabetical order followed by 4 distinct numbers in decreasing order
3) License plates with 2 distinct letters in alphabetical order followed by 4 distinct numbers
Can someone explain me how to count these objects ?
Thanks for your help.
Best Answer
It looks like these questions are really intended to hit at the distinction between permutations and combinations:
$\frac{n!}{(n-r)!}$ counts choices of $r$ objects from $n$ objects: you do not care about the order of the $r$ things (permutations).
$\frac{n!}{r!(n-r)!} = \binom{n}{r}$ counts choices of $r$ objects from $n$ objects: you care about the order of the $r$ things (combinations).
With this in mind, you can approach the problems like this:
The number of ways of choosing two distinct letters (and not caring about their alphabetical ordering) is $\frac{26!}{(26 -2)!} = 26\cdot25$. The number of ways of choosing four distinct digits in ascending (or descending) order is $\binom{10}{4}$. This means that there are $26\cdot25 \cdot \binom{10}{4}$ such licence plates.
Now you care about the order of both numbers and letters so you should count combinations for each, and then multiply the two.
You should count the possible combinations of two letters and the possible permutations of four numbers.