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.
I made this problem up (and its solution), so I guess I should answer the question, "Is it as simple as...". Yes, it's that simple, because (when n and m are unequal) eating exactly n meals and eating exactly m meals are independent events.
Please note, however, that this problem was based on a description of Feynman’s original that turned out to be inaccurate. In Feynman's version meals are rated on a continuum in the range [0,1], rather than being ranked 1 thru N, and the information available to the diner each night includes the rating P of the best meal tried so far, the number of nights n remaining to dine, and a threshold value Pn (that depends only on n) such that if P>Pn the diner repeats the meal rated P (“the best so far”), or otherwise s/he tries something new. Feynman found that to maximize the sum of the diner's meal ratings one should choose Pn = Sqrt[n]/(1+Sqrt[n]). For more detailed information see this page.
Michael A. Gottlieb
Editor, The Feynman Lectures on Physics New Millennium Edition
Best Answer
If the four meals are $A,B,C$, and $D$, we’re interested in the $4$-tuple $\langle a,b,c,d\rangle$, where $a$ is the number of $A$ meals, $b$ is the number of $B$ meals, and so on. Of course $a,b,c$, and $d$ must be non-negative, and $a+b+c+d=7$. The number of possible $4$-tuples is $$\binom{7+4-1}{4-1}=\binom{10}3=120\;,$$ as you already knew.
This is fine, if the cook chose one of these $4$-tuples at random with equal probability. If, on the other hand, he prepared a sequence of seven meals, choosing the type of each meal independently, then these $120$ combinations are not equally likely, and the probability of getting the right one is not $1/120$.
To see why these $120$ possible combinations of meals are not equally likely, pick one, $\langle a,b,c,d\rangle$. In how many different ways can the cook make this combination of meals? He makes them in some sequence. There are $\binom7a$ ways to choose which $a$ places in the sequence are $A$ meals. Then there are $\binom{7-a}b$ ways to choose which of the remaining $7-a$ places are $B$ meals. Continuing in the same vein, we see that there are $$\binom7a\binom{7-a}b\binom{7-a-b}c\binom{7-a-b-c}d\tag{1}$$ ways to make $a$ $A$ meals, $b$ $B$ meals, and so on.
Now simplify $(1)$: the last factor is $\binom{d}d=1$, so it’s $$\frac{7!}{a!(7-a)!}\cdot\frac{(7-a)!}{b!(7-a-b)!}\cdot\frac{(7-a-b)!}{c!(7-a-b-c)!}=\frac{7!}{a!b!c!d!}\;,$$ and you can easily verify that this multinomial coefficient depends on the specific values of $a,b,c$, and $d$. In particular, it’s $1$ when $a=7$ (or when $b,c$, or $d$ is $7$): by this procedure the cook is very unlikely to prepare $7$ identical meals. On the other hand, it’s $630$ when three of $a,b,c$, and $d$ are $2$ and the remaining one is $1$.