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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.

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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:

1) License plates with 2 distinct letters followed by 4 distinct numbers in increasing order

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.

2) License plates with 2 distinct letters in alphabetical order followed by 4 distinct numbers in decreasing order

Now you care about the order of both numbers and letters so you should count combinations for each, and then multiply the two.

3) License plates with 2 distinct letters in alphabetical order followed by 4 distinct numbers

You should count the possible combinations of two letters and the possible permutations of four numbers.

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  • $\begingroup$ Thanks for this detailed explanation. So, for 2) I counted $\binom{26}{2}\binom{10}{4}$ license plates and for 3) $\binom{26}{2} \frac{10!}{6!}$ license plates. $\endgroup$ – user513928 Jan 1 '18 at 21:18
  • $\begingroup$ @Sunny: No problem! Your answers look correct to me. $\endgroup$ – Alex Riley Jan 1 '18 at 21:24
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The key observation that we need to deal with the ordering constraints is that there are $\binom{n}{k}$ ways to choose $k$ distinct elements out of $n$, and that for each of those choices there is only one way to arrange the $k$ elements so that they are ordered. Hence $\binom{26}{2}$ counts the ways to pick two letters in alphabetic order, and $\binom{10}{4}$ counts the ways to pick four digits in descending order.

If the first digit is allowed to be $0$, $\binom{10}{4}$ is also the number of ways to pick four digits in ascending order. This assumption is likely to hold for license plates; otherwise, $10$ must be replaced by $9$.

The solution that you already know tells you how to deal with distinct, but not ordered elements. If letters and digits may come in any order, for every choice of two distinct letters and four distinct digits there are $6!$ ways to arrange them. When the letters come before the digits, we reason similarly, but we separately count the ways to fill the letter positions and the ways to fill the digit positions.

For example, if the letters only need to be distinct, while the digits must be in decreasing order, we get $\binom{26}{2} \cdot 2! \cdot \binom{10}{4}$. Do you see how to continue?

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