You can derive the probability in a manner similar to that for the usual derivation of the derangement probabilities (the probability that none of the $N$ men get their own hat back).
There are a total of $N!^n$ ways that all of the items can be distributed among the $N$ men so that each has exactly one of each type of item. Let $A_i$ denote the event that the $i$th man obtains the $n$ items that belong to him. The number of ways this can happen is $|A_i| = (N-1)!^n$, as this involves distributing all items but those belonging to the $i$th man among the other $N-1$ men. Similarly, $|A_i \cap A_j| = (N-2)!^n$, and, in general, $|A_{i_1} \cap A_{i_2} \cap \cdots \cap A_{i_j}| = (N-j)!^n$. There are also $\binom{N}{j}$ ways to choose which $j$ men will receive their own $n$ items.
Let $D(N,n,k)$ denote the number of ways that exactly $k$ of the $N$ men receive all $n$ of their items back. By the principle of inclusion/exclusion, $$D(N,n,0) = \sum_{j=0}^N (-1)^j \binom{N}{j} (N-j)!^n = N! \sum_{j=0}^N (-1)^j \frac{(N-j)!^{n-1}}{j!}.$$
Now, $D(N,n,k) = \binom{N}{k} D(N-k,n,0)$, as this is the number of ways of choosing $k$ of the $N$ men to receive all of their items back times the number of ways that none of the remaining $N-k$ men receive all of their items back.
Thus $$D(N,n,k) = \binom{N}{k} (N-k)! \sum_{j=0}^{N-k} (-1)^j \frac{(N-k-j)!^{n-1}}{j!} = \frac{N!}{k!} \sum_{j=0}^{N-k} (-1)^j \frac{(N-k-j)!^{n-1}}{j!}.$$
Dividing by $N!^n$, we have that the probability that exactly $k$ of the $N$ men receive all $n$ of their items back is
$$\frac{1}{k! N!^{n-1}} \sum_{j=0}^{N-k} (-1)^j \frac{(N-k-j)!^{n-1}}{j!}.$$
Note that this formula agrees with the values obtained by Henry for the case $N = 4$, $n=2$.
Added: In fact, the Poisson approximation suggested by Henry appears to match up well with the exact values provided by the formula given here for small values of $k$. The accuracy of the Poisson approximation appears to deteriorate, relatively speaking, as $k$ increases. However, the Poisson approach still gives a good absolute approximation when $k$ is large because the probabilities are extremely small.
Because many of your arrangements are mutually exclusive because they include the same unique article of clothing. If one man is wearing the big red hat, any combination using that article is invalid.
An alternate route to the answer more along the lines of your thinking would be to dress the first man $6\times5\times4$. Decrement each category by one. Dress the second man: $5\times4\times3$, then dress the third man: $4\times3\times2$. Multiplying those 3 terms will give you the correct answer.
Best Answer
If the coats are distributed at random so everybody has one coat then there are $n!$ possible distributions.
For (a) everybody having the wrong coat, you are looking at derangements and you have your expression, though this should be over a finite sum: $ \displaystyle d(n)=n!\sum_0^n \frac{(-1)^k}{k!}$ though I prefer to use $d(n)=\text{round}\left[\frac{n!}{e}\right]$. The probability is then $\frac{d(n)}{n!}$.
For (b) at least one person having the correct coat, this is the complement event of (a) and so very easy.
For (c) exactly one person having the correct coat, to count the possibilities you need to identify the individual with the correct coat and then count the derangements among the others, making the probability $\frac{n\,d(n-1)}{n!}$