## The SET® card game is played with a pack of $81$ distinct cards. Each card has four features (Shape, Color, Number, Shading). Each feature has three different variants (e.g. Color can be red, purple, green).

A SET consists of three different cards such that each feature is either the same on each card or different on each card.

For a collection $C_n$ of $n$ cards, let $S(C_n)$ denote the number of SETs in $C_n$. Then define $F(n) = \sum\limits_{C_n} S(C_n)^4$ where $C_n$ ranges through all collections of $n$ cards (among the $81$ cards).

You are given $F(3) = 1080$ and $F(6) = 159690960$.

Find $F(12)$.

$\scriptsize{\text{SET is a registered trademark of Cannei, LLC. All rights reserved.

Used with permission from PlayMonster, LLC.}}$

### To start, let’s understand that a type of SET depends completely on how many features are the same across all three cards. Since any given SET can either have 0, 1, 2, or 3 shared features across the three cards, we count the amount of different SETs for each case and sum:

0 shared features: Just 1 SET, since any combination of different features over three cards suffices. The order of the features does change the type of set. For example, suppose Card 1 has features A, B, C, D. Card 2 has features E, F, G, H. And Card 3 has features I, J, K, L. This yields a different type of set than if Card 1 had features E, F, G, H (features Card 2 had previously) while Card 2 now has features A, B, C, D.

1 shared feature: Choose 1 of the 4 features to keep constant: $C(4,1)$ ways. Then choose which of the 3 possiblities that feature will be: $C(3,1)$ ways. The remaining 3 features that change across the three cards can arrange in $3! = 6$ ways. This gives $C(4,1) * C(3,1) * 6 = 72$ sets.

2 shared features: Using similar logic, $C(4,2) * C(3,1)^2 * 2! = 108$ sets.

3 shared features: Again, $C(4,3) * C(3,1)^3 = 108$ sets.

Adding up these possibilities, we see that there are $1 + 72 + 108 + 108 = 289$ possible SETs.

Because there are $81$ cards, $C(81,3)$ ways to choose a SET, and $289$ possible SETs, the probability a random assortment of $3$ cards from the pile forms a SET is $\frac{289}{C(81,3)}$.

However we don’t stop there, as $F(3)$ includes repetitions. There are $C(81,3)$ amounts of $3$-collections, and if we choose any one of these collections, the probability that it is a SET is the value we found earlier. If it is a SET, the number of SETs is $1$. Thus, $S(C_3)$ for this specific $3$-collection is $1$, so the value of $S(C_3)^4$ is also $1$. Notice that there are exactly $289$ SETs, which means that $F(3) = 1+1+…+1$ $289$ times or $F(3) = 289$.

Before continuing, let’s revise $F(3)$, and convert it so that it is in the form of combinations and probabilities to mirror what was done earlier. Notice that the value of $F(3)$ is simply enumerating the amount of valid $3$-collections. This means that $F(3) = C(81,3)*\frac{289}{C(81,3)} = 289$. This seems pretty useless, but it will make sense why we want to do this when we move onto $F(6)$.

Notice that for a certain $6$-collection, there are $C(6,3)$ ways to choose $3$ cards from those $6$ cards. Each of these collections has a $\frac{289}{C(81,3)}$ chance to be a $3$-collection. But because order matters in this case, we also need to take into account how many ways we can arrange these sets.

To explain this, suppose we pick a $6$-collection. There are $C(6,3)$ ways to choose $3$ cards, or a $3$-collection, from this $6$-collection. Treated separately, each of these collections can choose from $289$ sets, since there are $289$ possible sets. So the total number of arrangements of sets in these collections separately is $C(6,3) * 289$.

However, we previously mentioned that the order of choosing these sets matters. This makes sense because if we have a $6$-collection and take out a $3$-collection, the second $3$-collection is completely determined since there are only $3$ cards left in the $6$-collection. Thus, the order of which $3$-collection we choose actually matters.

As a result, the total number of ways to arrange sets in the $6$-collection is $2!*C(6,3)*289 = 2*C(6,3)*289$. However, we’ve ignored the $S(C_{6})^4$ part of $F(6)$, meaning that we actually have to raise the value we obtained to the fourth power. Thus, we obtain a total number of arrangements to be $(2*C(6,3)*289)^4$

We have found the value of $S(C_{6})^4$ for a specific $6$-collection. So we can multiply it by the total number of collections, which is $C(81,6)$, and sum this product from $1$ to $C(81,6)$ to obtain $F(6)$. To explain this, if we choose a specific $6$-collection, there are $C(81,6)$ ways of doing so. For that specific $6$-collection, the amount of valid set arrangements is $(2*C(6,3)*289)^4$. Thus, there are a total of $C(81,6)*[2*C(6,3)*289]^4$ set arrangements for the $6$-collection.

Following this procedure, there are $C(81,6)*[\frac{289}{C(81,3)}*2*C(6,3)]^4 = 159690960$, which we remember is $F(6)$. With this procedure, we can repeat our steps to solve for $F(12)$.

In a $12$-collection, the ways to choose $3$ cards is $C(12,3)$, hence there are $C(12,3)$ $3$-collections. In those $3$-collections, there are $289$ ways to arrange the set. However, because the order of choosing these SETS matters, the amount of ways to arrange these $12$-collection is just $(C(12,3)*4 * 289)^4$. Following our previous procedure, we multiply this value by the total amount of $12$-collections, which is $C(81,12)$, to obtain $F(12) = C(81,12)*[\frac{289}{C(81,3)} * 4 * C(12,3)]^4$.

After calculating, we find $F(12) = \boxed{2060386807998775365274204000}$.

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