# Combinatorics - small lottery with different outcomes

The question goes like this:

Given a small lottery of 8 people, how many possible ways can the lottery's prizes be handed out in the following scenarios: Scenario A - there are 8 different prizes, and exactly one person does not win a prize? (One person wins two prizes, and the order does not matter). Scenario B - there are 8 identical prizes and exactly one person does not win a prize? (One person wins 2 out of the 8 identical prizes).

So far I thought a possible way to find an answer to A would be to use $$8P7 * 7$$ because we would start choosing from 8 people, then 7 and so on until we have chosen 7 winners. Then out of the 7 winners, we choose 1 to receive the extra prize. This has given me 282240 different ways on doing it, which apparently is incorrect. How would I go about solving this kind of problem?

There are eight ways to select the person who receives two prizes, $$\binom{8}{2}$$ ways to select two of the eight prizes for that person, $$\binom{7}{6}$$ ways to select which six of the remaining seven people receive a prize, and $$6!$$ ways to distribute the remaining six prizes to those six people so that each person receives one of those prizes. Hence, there are $$\binom{8}{1}\binom{8}{2}\binom{7}{6}6!$$ ways to distribute the eight prizes to eight people so that exactly one person does not receive a prize.