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!'.
$$ 5! ({20 \choose 5} - {18 \choose 3}) $$
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 $5!\times(\binom{18}{5}+2\binom{18}{4})$.