2024 SSMO Accuracy Round Problems/Problem 9

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Problem

In the game of $S$ - set, there are $3^{12}$ unique cards, each card containing a 12 traits of variants $A,B,$ and $C.$ A full house in the game of Sset is defined to be a hand of cards in which for each trait, all cards in the hand either share the same variant or have all different variants. The number of hands that can be considered a full house in the game of set can be expressed as $a^b+c^d+e^f,$ where $a,b,c,d,e,$ and $f$ are positive integers and $a+c+e$ is minimized. Find $a+b+c+d+e+f.$

Solution

Note that all full houses can contain at most $3$ cards. Now, 2 cards uniquely determine a hand. So, there are $\binom{3^{12}}{2}$ full house hands of 2 cards. As there are 6 permutations for 3 card hands, there are $\frac{(3^{12})(3^{12}-1)}{6}$ full house hands of 3 cards. Lastly, note that all one-card hands are considered full houses. Thus, the answer is \[\frac{3^{24}-3^{12}}{2}+\frac{3^{24}-3^{12}}{6}+3^{12} = \frac{2}{3}(3^{24})+\frac{1}{3}(3^{12}) =\]\[3^{23}+3^{23}+3^{11}\implies 3+23+3+23+3+11 = \boxed{66}.\]

~SMO_Team