So I've got the problem "How many 5-letter words can be made if the alphabet has 29 letters and a) the letter A doesn't show once and b) the letter A shows at least once.
For A) I counted it as 28!/23! since you only have 28 letters to really think about without A? But I'm really stuck with B, so if anyone could help me, that would be appreciated!
If no A is used, there are $29 - 1 = 28$ ways to fill each position, so there are $28^5$ possible five-letter words that do not contain an A.
Your answer $$P(28, 5) = \frac{28!}{(28 - 5)!} = \frac{28!}{23!}$$ would be correct if letters could not be repeated.
The set of five-letter words in which the letter A is used at least once is the complement of the set of five-letter words in which the letter A is not used once. To find the answer, subtract the number of five-letter words in which the letter A is not used once from the number of five-letter words that can be formed from the alphabet.