# probability with and without repeats

The standard license plates take 3 letters followed by four digits. ex. ABC 1234 Assume that all plates are non-controversial and meet state restrictions (although we know this is not the case0.

a. how many license plates are possible without repetition of letters or numbers?

b. the standard issue license will repeat letters and/or numbers in the same style (3 letters followed by 4 numbers. How many standard issue plates are available.

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So for starter, look at how many choices there are for the first letter, 26. Then we have chosen one, so there are 25 choices for the second, and 24 for the third. From there we consider the numbers. There are 10 choices for the first number (because we are including 0) then there are 9 choices for the second because we used one. Then 8 and 7 for the third and fourth. Then multiply these together for $26 \cdot 25 \cdot 24 \cdot 10 \cdot 9 \cdot 8 \cdot 7$
Then for the repeat. Since we can repeat, we have 26 letter choices for each letter slot and 10 number choices for each number slot so we have $26 \cdot 26 \cdot 26 \cdot 10 \cdot 10 \cdot 10 \cdot 10$ or $26^310^4$
So you have an alphabet of $26$ letters, and there are $$\frac{26!}{23!}=26\cdot25\cdot24=25\cdot624=15600$$ ways to choose three letters (accounting for order) with no repetitions. For the digits, you have something very similar: $$\frac{10!}{6!}=10\cdot9\cdot8\cdot7=5040$$ so the answer to part a will be the product of these two independent sub-choices.
Part b is much easier: $26^3\cdot10^4$, since each letter and digit is completely independent of the others.