Multiset Counting Story Problem Ten friends check their hats with the coatroom clerk. Five are wearing fedoras and five are wearing berets. In how many ways can the clerk return the hats so that no one gets their own hat back if...
$a$. everyone gets the right kind of hat?
$b$. noboby gets the right kind of hat?
$c$. the clerk keeps the fedoras and only returns the 5 berets. In this scenario the five friends that do get a hat back may or may not get the right kind of hat.
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I at least believe this is utilizing the application of multisets to properly answer the question, I want to be sure though as this specific question is worth a lot of points! I appreciate any answers/tips
 A: I do not see multisets involved here, but I do not know if another look at this question may use them. Some hints:
a. Let's have a look at the fedora guys: We have five objects (the fedoras) and want to permute them without any fixed point. The number of ways in which that is possible is called the dearrangement number (a fixed point free permutation is called dearrangement): We have 
$$ d_5 = \mathord!5 = 44
 $$
Same for the berret guys. 
b. This is easier than a., five objects have to be distributed between five people (as nobody gets the right kind of hat, we can permute the berrets between the fedora guys), this is possible in 
$$ 5! = 120 $$
ways. Same for the fedoras.
c. One way to approach this is inclusion-exclusion. There are $\binom{10}5$ ways to choose the people that get a hat. We may distribute that hats in $5!$ ways, so there are 
$$ \binom{10}5 5! $$
ways to distribute the hats. But stop! Same people may have gotten their own hat. There are five people for which this is possible, one of them chosen ($5$ ways), there remain $\binom 944!$ ways to distribute the remaining hats, 
we have 
$$ \binom{10}55! - 5\binom 944! $$
ways. But, we have double counted the distributions where two people get their own hats, adding them again, gives $\binom 52$ ways to choose the two people and $\binom 833!$ to distribute the remaining hats 
$$ \binom{10}55! - 5\binom 944! + \binom 52 \binom 833! $$
But now, we have not correctly accounted for the distributions where three guys have their own hats, so we continue 
$$ \binom{10}55! - 5\binom 944! + \binom 52 \binom 833! - \binom 53\binom 722! $$
I'm sure, you can continue now.
