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 $1$.
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 we care about which person eats which meal and who drinks which beverage, the oldest person can choose a meal and a drink in $4 \cdot 7$ ways, the next oldest person can choose a meal and a drink from the remaining options in $3 \cdot 6$ ways, and the youngest person can choose a meal and a drink from the remaining options in $2 \cdot 5$ ways. Hence, if we care who orders which meal and who orders which drink, the three people can place $$4 \cdot 7 \cdot 3 \cdot 6 \cdot 2 \cdot 5$$ orders. This is what you have calculated.
If you meant the order in which the meals and drinks are ordered does not matter, then your answer is correct. If you meant it does not matter which person orders which meal or which person orders which drink, read the next paragraph.
If we only care about which meals and which drinks are consumed, then the three meals can be selected from the four available meals in $\binom{4}{3}$ ways and the three drinks can be selected from the seven available drinks in $\binom{7}{3}$ ways. Hence, there are $$\binom{4}{3}\binom{7}{3}$$ ways to select three different meals and three different drinks at the restaurant.