Sushi Go Party! includes a lot of options not available in the original Sushi Go. Players can select what on the menu\footnote{Unfortunately, there's also a menu item titled menu.} for each play. A valid menu has a set format: Nigiri is always included, as well as one of three types of roll, three out of eight types of appetizer, two out of eight types of specials, and one of three types of dessert. Each of these choices is independent, so we can multiply the number of ways to do each together to get the total number of combinations. For appetizers and specials where we're choosing more than one, we don't care about the order, what matters is what items are shuffled into the deck. Thus, we're looking for the number of combinations of each, where the number of combinations of selecting $k$ items from $n$ options, is

\begin{align} \binom{n}{k} = \frac{n!}{k! \cdot (n-k)!}. \end{align}If you want to be able to calculate this manually without a tool with a special function, it can be helpful to do some of the cancellation yourself. We can rewrite this while reducing the amount of common terms in the numerator and the denominator.

\begin{align} \frac{n!}{k! \cdot (n-k)!} = \frac{\prod_{i=n-k+1}^n i}{k!} \end{align}A concrete example may help.

\begin{align} \binom{8}{3} &= \frac{8!}{3! \cdot (8-3)!}\\ &= \require{cancel}\frac{8 \cdot 7 \cdot 6 \cdot \cancel{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}}{3 \cdot 2 \cdot 1 \cdot \cancel{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}} \\ &= 56 \end{align}Applying this we get the number of combinations.

\begin{align} 1 \cdot 3 \cdot \binom{8}{3} \cdot \binom{8}{2} \cdot 3 = 14112 \end{align}There are a couple of wrinkles. First, is that a couple of the menu items score a little bit differently depending on how many players there are, but I don't consider that a different menu.

Second, the above only applies for player counts of 3--6. When there are 2, 7, or 8 players there are some restrictions to the items included. Spoon (a special) and edamame (an appetizer) may not be used in two-player games. Thus in such cases there are fewer menu combinations.

\begin{align} 1 \cdot 3 \cdot \binom{7}{3} \cdot \binom{7}{2} \cdot 3 = 6615 \end{align}Similarly, in seven- and eight-player games neither menu nor special order are allowed, both of which are specials.

\begin{align} 1 \cdot 3 \cdot \binom{8}{3} \cdot \binom{6}{2} \cdot 3 = 7560 \end{align}
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