Ordering Dishes: Probability and Combinations I'm struggling with the following question:
A restaurant has $5$ dishes, if a group of four orders from the restaurant what is the probability that $2$ people order the same dish and the $2$ others order different dishes from each other and the $2$ that ordered the same. (ex AABC)
My work so far is:
Total number of ways to order 4 dishes from $5 = \frac{8!}{4!4!}$
I'm having trouble figuring out how many combinations fit my criteria. Any help or hints would be appreciated.
 A: 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
$$P = \frac{n}{N} = \frac{6\times 5\times 4\times 3}{5^4} = \frac{72}{125}\,.$$
