If letters and numerals can be repeated, it will in fact be $36^6$. For each of the six slots, we have 36 choices for what can go there. By the rule of product, there are then $36\cdot 36\cdot 36\cdot 36\cdot 36\cdot 36=36^6$ different such license plates.
If letters and numerals cannot be repeated, you will have 36 choices for the first slot, 35 choices for the second, 34 for the third,... for a total of $36\cdot 35\cdot 34\cdot 33\cdot 32\cdot 31 = \frac{36!}{30!}$
I'll leave the final one to you to think about, but again, try to break it up via rule of product by figuring out how many choices there are for the first spot, how many choices for the second, how many choices for the third, etc... and multiply those numbers together.
No. At the very least, you also need to know the a priori probability that a plate is a vanity plate. That is, when you ask:
is there a way to determine the probability that a license plate with no digits was randomly chosen and is not a vanity plate?
you are asking for the conditional probability $P(V|N)$ with $V$ being the event of the plate being a vanity plate, and $N$ being the event of the plate having no digits.
Now, you could of course try to use Bayes' formula:
$$P(V|N)=\frac{P(N|V)\cdot P(V)}{P(N)}=\frac{1\cdot P(V)}{P(N)}= \frac{ P(V)}{P(N)}$$
But, like I said, you'd need to know the a priori probability $P(V)$ to use this.
Moreover, the $P(N)$ that this formula refers to is not the same as the $P(N)$ that you calculated: the $P(N)$ you calculated was the probability of getting no digits on a plate where all symbols are picked randomly, i.e. this was assuming no one has requested a vanity plate!
Now, here is something you can do: find the actual $P(N)$, i.e. find the percentage of plates that are actually out there being used that have no numbers. If you find that this is far more than that $5$% that you would get when creating plates randomly, then you have good reason to believe that lots of people got vanity plates. If, however, you find that this $P(N)$ is close to that $5$%, then that suggest not many people order vanity plates. So .. start counting! :)
Best Answer
That is incorrect.
First, let's find the number of license tags, total, that can be made with $3$ letters and $3$ digits. This is: $$26\cdot26\cdot26\cdot10\cdot10\cdot10 = 26^310^3$$
Second, look at how many tags can be made without repeating any character (letter or number): $$26\cdot25\cdot24\cdot10\cdot9\cdot8$$
Now, if we must repeat at least one character, then number of tags that satisfy are: $$\text{Total} - \text{ThoseThatDon'tRepeat} = 26^310^3-26\cdot25\cdot24\cdot10\cdot9\cdot8$$