Is there < $2^n$ combinations of $length(S)/3$ when solving this special case of Exact Three Cover?

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Suppose I want to find an Exact Three Cover for $S$=$[1,2,3,4,5,6]$.

  • Length of $S$ must total to an amount divisible by $3$.

$K$ = $length(S)/3$

It requires $K$, $3$ $element$ $sets$ to generate an Exact Three Cover for $S$. (eg. $(1,2,3)$,$(4,5,6)$)

  • Given all possible combinations of $3$ for $S$, I create a list called $C$.

  • Take note that this list is polynomial because it is fixed to $3$.

Considerations for $C$

  1. $(1,2,3)$ is the same as $(1,3,2)$. A sorting algorithm would address this issue for solving Exact Three Cover. (eg. Delete (1,3,2), but leave (1,2,3))
  2. Repeating sets can trivially be removed.(of course, leave one of the sets)
  3. Sets with repeating elements or sets with elements that do not exist in $S$ are deleted by a sorting algorithm.
  4. Because of the sorting algorithm, the worst-case input for $C$ will always have a length equal to all possible combinations of 3 from $S$ (3-element sets)
  5. Almost forgot, no sets with elements that total $<$ $3$ or $>$ $3$.

Question

Is there < $2^n$ combinations of $K$ to choose from a list created from all possible combinations of 3-element sets?

Edit: $n$ represents the total amount of 3-element sets of $C$. (Which is all possible combinations of 3 from $S$)