Decision Problem: Given a set $ S$ , is there at least a given $ N$ $ >$ $ 1$ amount of solutions, for an $ Exact~Cover~by~3sets$ for $ C%$ ?
$ s$ = $ 1,2,3,4,5,6$
$ c$ = $ [[1,2,3],[4,3,2],[4,5,6],[5,1,6],[5,6,3]]$
Solutions
$ [1,2,3],[4,5,6]$
$ [4,3,2],[5,1,6]$
$ N$ = $ 2$
Yes, there are $ N$ solutions.
Algorithm

Remove sets that have repeating elements
(eg. [1,1,2] is deleted from $ C$ )

Remove sets that have elements that don’t exist in $ S$
(eg. [9,5,6] is deleted because $ 9$ not in $ S$ )

Make sure all elements in $ S$ exist in $ C$ .
$ for$ a $ in$ $ range(0, length(s)):$
$ ~~~~~~~~$ $ IF$ $ s[a]$ $ not$ in $ c$ :
$ ~~~~~~~~~~~~~$ OUTPUT NO
Convert $ C$ into a complete list
$ WHILE$ $ c[i]$ has [brackets]:
$ ~~~~~~~~~~$ DELETE [BRACKETS] FROM $ C$
now $ c$ = $ [1, 2, 3, 4, 3, 2, 4, 5, 6, 5, 1, 6, 5, 6, 3]$
Finally, Decide
$ n$ = $ (‘Enter~for~N:~’))$
$ yes$ = $ 0$
$ for$ a $ in$ $ range$ (0, $ length(c)):$
$ ~~~~~~$ $ if$ $ c$ .count($ c$ [a]) >= $ n$ :
$ ~~~~~~~~~~$ $ yes$ = $ 1$
$ ~~~~~~$ else:
$ ~~~~~~~~~~$ OUTPUT NO
$ ~~~~~~~~~~$ HALT
$ if$ $ yes$ == $ 1$ :
$ ~~~~$ OUTPUT YES
Edit: The above should do the same below.
yes = 0 for a in range(0, length(s)): if c.count(s[a]) >= n: yes = 1 else: OUTPUT('No') break if yes == 1: OUTPUT('yes')
Facts to consider

There cannot be any sets with elements that don’t exist in $ S$ .

There cannot be any sets with repeating elements.
 All elements in $ S$ must exist in $ C$ . Else, a $ no$ is given.
 $ N$ must be > $ 1$
 If any element in $ C$ occurs < $ N$ times then the output must be $ No$ , because there wouldn’t be at least $ N$ solutions.
Question
Will this algorithm always work if the input is > $ 1$ , and if no how would it fail?