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Sorry if this is super simple, I've just been stuck on this concept for way too long.

Each person can only win one prize.
How many ways can 5 prizes be handed out to a group of 20 people if:

(a) all 5 prizes are identical, also Alex and Sally can't both get prizes.

(b) the 5 prizes are ranked 1st-5th, also Alex and Sally can't both get prizes.

So I think I've tried to solve this but I think I've made a mistake.

New Attempt

a)
Now I'm trying to find the total number of ways that the 5 prizes could be handed out to 20 people. I think this should be: $$ {20 \choose 5} $$
now we need to find the situations in which both Sally and Alex win prizes and subtract that from the total. I think this is wrong but it might just be: $$ {18 \choose 3} $$ So the final value would be: $$ {20 \choose 5} - {18 \choose 3} $$

b)
I think the trick here is still to multiply the answer from part (a) by '5!'.
$$ 5! ({20 \choose 5} - {18 \choose 3}) $$

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  • $\begingroup$ Can a person receive multiple prizes, e.g. could Alex hypothetically get all five? $\endgroup$ – Eevee Trainer May 7 at 2:22
  • $\begingroup$ no, one apiece. $\endgroup$ – JBob May 7 at 2:34
  • $\begingroup$ If there are five prizes, why are you choosing six objects? For the first problem, subtract the number of ways Alex and Sally can both receive prizes from the total number of ways prizes can be distributed. $\endgroup$ – N. F. Taussig May 7 at 2:36
  • $\begingroup$ Can you explain the thought behind your proposed answers? $\endgroup$ – Henning Makholm May 7 at 2:37
  • $\begingroup$ sorry, choosing 6 objects was a typo on my part. I'll try to think of a way to find the situations in which both Alex and Sally receive prizes. I added my thoughts for my current work as well. $\endgroup$ – JBob May 7 at 2:52
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For a. possibilities are neither Alex or Sally get a prize $\binom{18}{5}$ or one of them gets a prize and the other doesn't $2\binom{18}{4}$, so total is $\binom{18}{5}+2\binom{18}{4}$.

For b. Use a. result and multiply by permutations - net $5!\times(\binom{18}{5}+2\binom{18}{4})$.

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  • $\begingroup$ Note: $\binom{18}{5}+2\binom{18}{4}=\binom{20}{5}-\binom{18}{3}$. $\endgroup$ – herb steinberg May 7 at 17:13

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