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
Rephrasing the question into a more common form, you are asking for the number of integral solutions to the following system:
$$\begin{cases} x_1+x_2+x_3 = 6\\ 0\leq x_1\leq 3 \\ 0\leq x_2\leq 4 \\ 0\leq x_3\leq 5\end{cases}\\\text{where each of}~~x_1,x_2,x_3\in\mathbb{Z}$$
You correctly noted above that if there was not a maximum limit above that the solution would be $\binom{6+3-1}{6}$, as per the stars-and-bars method.
We will couple this knowledge with the principle of inclusion-exclusion. The combinations we are interested in will be the set of all possible combinations without restriction minus those combinations which violate one or more of the restrictions.
Let $A_1$ be the set of all combinations such that you violate the upper bound on $x_1$ (i.e. you have strictly more than 3 chicken dishes). How many combinations exist in $A_1$ that we will need to subtract from the total?
Well, since the upperbound condition is violated, that means $4\leq x_1$. By a change of variable $y_1=x_1-4$, we arrive at the system:
$$\begin{cases} y_1+x_2+x_3 = 2\\ 0\leq y_1 \\ 0\leq x_2 \\ 0\leq x_3\end{cases}\\\text{where each of}~~y_1,x_2,x_3\in\mathbb{Z}$$
The number of which then is $\binom{2+3-1}{2}$.
Continue calculating $A_2$ and $A_3$ which will represent having violated $x_2$'s upper bound and $x_3$'s upper bound respectively (i.e. having sold too many beef and lamb dishes).
In general though, it is possible for these to intersect (that you could have simultaneously sold too many chicken and simultaneously sold too many lamb dishes). You will need then to calculate $A_{1,2}, A_{1,3}, A_{2,3}$ and possibly even $A_{1,2,3}$ denoting having violated each of the corresponding conditions simultaneously.
Letting $B$ denote no violated conditions, and $S$ denoting no conditions present in the first place, by principle of inclusion-exclusion the answer will be:
$|B| = |S| - |A_1| - |A_2| - |A_3| + |A_{1,2}| + |A_{1,3}| + |A_{2,3}| - |A_{1,2,3}|$
Which in this case is:
$=\binom{6+3-1}{6} - \binom{2+3-1}{2} - \binom{1+3-1}{1} - \binom{0+3-1}{0} + 0 +0 + 0 - 0$
(noting that it is impossible in this case to simultaneously sell too many of two types of dish at the same time)
To generalize this problem to the larger case, let $A_{i}$ denote violating the $i^{th}$ condition (and possibly more), $A_{i,j}$ denote violating the $i^{th}$ and $j^{th}$ conditions, ... $A_{i,j,\dots,p}$ denote violating each of the conditions $i$ through $p$, and even more generally for an index set $\Delta \subset \{1,2,\dots,k\}$ you will have $A_\Delta$ violating all of the conditions on $x_i$ for each $i\in\Delta$:
$$|B| = |S| + \sum\limits_{i=1}^k\left((-1)^i\sum\limits_{|\Delta|=i}|A_{\Delta}|\right)$$
And $|A_\Delta| = \binom{n +k - 1- \sum\limits_{i\in\Delta}(r_i+1)}{n}$, where $r_i$ is the upper bound for $x_i$
Best Answer
The total number of different possibilities is given by considering that each one of the four people can choose any of the five dishes, independently from the other costumers. So $N=5^4$.
Now the number of choices you're interested in can be evaluated as follows. Let's label the four people as $A$, $B$, $C$, $D$. There are are $6= {4\choose 2}$ different pairs (the two people taking the same dish), explicitly $(A,B)$, $(A,C)$, $(A,D)$, $(B,C)$, $(B,D)$, $(C,D)$.
Now consider that the couple is $(A,B)$. Then $A$ has $5$ possibilities of choice. Once $A$ has chosen, the choice of $B$ is forced, since it's the same choice of $A$. Then $C$ has to choose one among $4$ available dishes and finally $D$ has $3$ possibilities of choice. So if the pair $(A,B)$ is the one taking the same dish, then there are $5\times 4\times 3$ possibilities.
To have the total number of possibilities with this sort of configuration we multiply by $6$, the number of possible pairs.
So the configurations are $n=6\times 5\times 4\times 3$.
Finally the probability is $$P = \frac{n}{N} = \frac{6\times 5\times 4\times 3}{5^4} = \frac{72}{125}\,.$$