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.

  • 2
    $\begingroup$ There are presumably $5^4$ equally probable order patterns, which you might consider could come as (a) all four the same, (b) all four different, (c) three the same and one different, (d) two pairs, and (e) two the same and the other two individually different. You want to find the number of type (e) orders. $8!$ is not part of the calculation $\endgroup$
    – Henry
    Aug 10, 2020 at 9:36
  • $\begingroup$ Your approach to finding total number of ways to order $4$ dishes is not correct. What you are finding will only arrange $4$ out of $5$ dishes in $4$ places in various ways - this will include cases where one person could get all $4$ dishes and other $3$ could go hungry. The right way to look at it will be that each of them have $5$ independent choices so in total, there are $5^4$ possibilities. $\endgroup$
    – Math Lover
    Aug 10, 2020 at 9:50
  • $\begingroup$ $\binom{8}{4}= 70$ not 5 (!) $\endgroup$ Aug 10, 2020 at 9:52
  • $\begingroup$ @Henry there are actually $70$ order patterns $\endgroup$ Aug 10, 2020 at 9:54
  • $\begingroup$ @HennoBrandsma - Do you think those 70 are equally likely? Suppose the restaurant had 2 dishes and 2 customers. Would you say the probability they order different dishes is $\frac13$? $\endgroup$
    – Henry
    Aug 10, 2020 at 10:01

1 Answer 1


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 $$P = \frac{n}{N} = \frac{6\times 5\times 4\times 3}{5^4} = \frac{72}{125}\,.$$

  • $\begingroup$ How do you take into account that the orders are considered the same regardless of who orders what exactly? The number of dishes and which ones, is all that counts. So (if dishes are A to E, AABCD is the same as BCDAA etc.) $\endgroup$ Aug 10, 2020 at 9:56
  • $\begingroup$ order AABC is counted several times, as AABC and AACB under pair 1,2; and BAAC and CAAB under pair 2,3 etc. How do you account for that? $\endgroup$ Aug 10, 2020 at 10:23

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