Let $S$ be the universal set, where:
\[
S=\{1,2,3, \ldots, 18,19,20\}
\]

Let sets $A$ and $B$ be subsets of $S$, where:

Set $A=\{2,10,11,14,15,19\}$

Set $B=\{1,4,5,6,7,9,11,13,15,17,19\}$
Set $C=\{1,4,8,12,17,18,20\}$

Find the cardinality of the set $(A \cap C) \cap B^{c}$
\[
n\left[(A \cap C) \cap B^{c}\right]=
\]

You may want to draw a Venn Diagram to help answer this question.

Step 1 ：Find the intersection of sets A and C, denoted as \(A \cap C\). Given that \(A = \{2,10,11,14,15,19\}\) and \(C = \{1,4,8,12,17,18,20\}\), we find that \(A \cap C = \{ \}\) as there are no common elements between A and C.

Step 2 ：Find the complement of set B in the universal set S, denoted as \(B^{c}\). Given that \(S = \{1,2,3, \ldots, 18,19,20\}\) and \(B = \{1,4,5,6,7,9,11,13,15,17,19\}\), we find that \(B^{c} = S - B = \{2,3,8,10,12,14,16,18,20\}\).

Step 3 ：Find the intersection of the set \(A \cap C\) and the set \(B^{c}\), denoted as \((A \cap C) \cap B^{c}\). Given that \(A \cap C = \{ \}\) and \(B^{c} = \{2,3,8,10,12,14,16,18,20\}\), we find that \((A \cap C) \cap B^{c} = \{ \}\) as there are no common elements between \(A \cap C\) and \(B^{c}\).

Step 4 ：Therefore, the cardinality of the set \((A \cap C) \cap B^{c}\), denoted as \(n\left[(A \cap C) \cap B^{c}\right]\), is 0. Hence, \(\boxed{0}\) is the solution to the problem.