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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?

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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.

Your approach did not take into account which person received which prize(s).

Addendum: If the prizes are identical, select which of the eight people receives two prizes and which six of the other seven people each receive one prize.

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  • $\begingroup$ Thank you, this appears to be a correct answer to A. How would you go about solving B? Assuming the prizes are identical, should the different ways to select two of the eight prizes for a specified person be different? $\endgroup$ Commented May 4, 2022 at 12:53

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