Probabilities on giving mixed items correctly I am stuck with the following problem and any help would be appreciated.
A group of $9$ goes to a restaurant.
Two persons order steaks, $3$ order burgers and $4$ order pizzas.
The waiter forgets who ordered what.
What is the probability when bringing the food to the table, that everyone gets the food they ordered?
Edit: my approach to this problem which I presume is wrong is that each stake person has a $2/9$ probability to get the right food, each burger person $3/9$ and each pizza person $4/9$ and then multiply these fractions, but the result is so counter-intuitive small that I truly believe I have messed something up.
 A: Assuming we have $M=3$ different meals for $P=9$ people, the probability of everyone getting what they ordered can be thought as the ratio between the number of correct cases and the total number of possible cases:
$$
p=\frac{N_{ok}}{N_{tot}}
$$
From the generic orders per person, we know the sets of people $P_i$ that have ordered the $i$-th meal, we have the total number of possible correct meal combinations as:
$$
N_{ok}=\prod_{i=1}^M P_i!=2!\cdot 3!\cdot 4!=288
$$ 
this number includes mixed possibilities: if Alice and Bob order each one a different pizza, then $P_{pizza}=2$ and there are $2!=2$ mixed possibilities, that either Alice and Bob get what they ordered or they will get their pizzas swapped. This holds true if Alice and Bob are very hungry and willing to eat pizza, if everyone wants exactly what they ordered then only one configuration will be accepted, or $N_{ok}=1$. 
The number of total possible combinations of $M$ meals for $P$ people, with repetitions (so including the case all people get the same meal) is:
$$
N_{tot}=M^P=3^9=19683
$$
That is, assumed that the waiter totally forgot the order, but remembered that the group ordered 3 meals. If the waiter just forgot the names (and so, the number of each meals is known), then:
$$
N_{tot}=P!=9!=362880
$$
The probability of the waiter to tell the chef the correct meal to this set of people will be approximately $1.463\%$ in the former case, and $0.079\%$ in the latter case.
A: Your intuition is good. I'm assuming that the meals are indistinguishable (that is, every burger, steak and pizza is the same). 
Another (more combinatorial way) to approach this problem is as follows. Let $S$ stand for steak, $B$ stand for Burger, and $P$ stand for pizza. Without loss of generality, you can assume that the people are all sitting in a single line at the same table (imagine they ordered at a bar or something) and that the the orders came in as $O = S,S,B,B,B,P,P,P,P$. 
There are 9! ways for the people to receive their orders, of these 9! distinct permutations of these letters you have to determine how many are equivalent to $O$. there are 2! ways to arrange the first two $S$'s, 3! ways to order the $B$'s and 4! ways to order the $P's$ so that they all wind up contingent.  
Thiss means that there are exactly 
$$N=2! \cdot 3! \cdot 4!$$
permutations of these letters that are equivalent to $O$.
Since there are 9! possible orderings of these letters, the probability that the waiter gets the order correct is 
$$\frac{N}{9!} = \frac{1}{1260}$$
This is actually different from the other answer provided, I think that there was a mistake in assuming that there are $3^9$ ways that the waiter could return the orders. this number ($3^9$) counts every possible string of $S$'s, $B$'s and $P$'s, but we only want strings with exactly 2 $S$'s, 3 $B$'s and 4 $P$'s. 
Hope this helps
