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I am trying to calculate the total number of possible orders from a group of $4$ friends at a restaurant and there are $12$ different dishes to choose from. Each friends has to have a different dish from each other. So far, I have this calculation:

$12 * 11 * 10 * 9 = 11,880$

Does this look right?

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    $\begingroup$ Yes, it does. Assuming that an "order" reflects the individual choices. Thus, you are assuming that $ABCD$ is different then $ABDC$ even though someone else might say that in each case the "order" given to the server is "one each of $A,B,C$ and $D$" I'd say the phrasing of the question was ambiguous, myself. $\endgroup$
    – lulu
    Jan 22, 2022 at 17:20
  • $\begingroup$ Shorthand notation is $~\displaystyle \frac{n!}{(n-k)!}~$ where $~(n,k) = (12,4).~$ Compare this with $~\displaystyle \binom{n}{k} = \frac{n!}{(n-k)! \times k!}~$ which represents the number of ways that $k$ distinct orders could be chosen from $n$ orders, without any regard to which of the $k$ people got which orders. $\endgroup$
    – user2661923
    Jan 22, 2022 at 18:14

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This looks right! Another way to think about this is the first friend orders and has $12$ choices. The second friend can then order anything that the first friend didn't, so they have $11$ choices. The third friend similarly has $10$ choices and the final friend has $9$ choices.

So by the multiplication principle, the total number of orders is $12 \ast 11 \ast 10 \ast 9$ as you stated.

As someone mentioned in a comment, it is worthwhile to note this is different from asking the question: "What are all the possible ways to choose $4$ distinct dishes from the $12$ possible dishes?" Since in your situation, we do care which friend gets which meal.

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The formal method to calculate this is to use permutation notation. In this case we have $n = 12$ dishes to choose from and $k = 4$ people choosing things. We also have that they cannot chose the same dish. So we use the formula

\begin{align} \frac{n!}{(n - k)!} &= \frac{12!}{(12 - 4)!}\\ &= \frac{12!}{8!}\\ &= \frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8!}{8!}\\ &= 12 \cdot 11 \cdot 10 \cdot 9 = 11,880 \end{align} so you are correct!

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  • $\begingroup$ I don’t think this is “choose notation”? What you’re describing is permutation, not combination. $\endgroup$
    – VTand
    Feb 9 at 5:11
  • $\begingroup$ Good catch, I've edited it. $\endgroup$
    – Clyde Kertzer
    Feb 9 at 5:17

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