Suppose $ A_1,\dots, A_6$ are six sets each with four elements and $ B_1,\dots,B_n$ are $ n$ sets each with two elements, Let $ S \equal{} A_1 \cup A_2 \cup \cdots \cup A_6 \equal{} B_1 \cup \cdots \cup B_n$. Given that each elements of $ S$ belogs to exactly four of the $ A$'s and to exactly three of the $ B$'s, find $ n$.
1. **Understanding the Problem:**
   We are given six sets \( A_1, A_2, \ldots, A_6 \) each with four elements, and \( n \) sets \( B_1, B_2, \ldots, B_n \) each with two elements. The union of all \( A_i \)'s is equal to the union of all \( B_i \)'s, denoted by \( S \). Each element of \( S \) belongs to exactly four of the \( A \)'s and exactly three of the \( B \)'s. We need to find \( n \).

2. **Cardinality of \( S \):**
   Let \( |S| = m \). Each element of \( S \) appears in exactly four of the \( A_i \)'s. Since each \( A_i \) has four elements, the total number of appearances of elements in all \( A_i \)'s is \( 6 \times 4 = 24 \). Therefore, we have:
   \[
   4m = 24 \implies m = 6
   \]
   So, \( |S| = 6 \).

3. **Counting Elements in \( B_i \)'s:**
   Each element of \( S \) appears in exactly three of the \( B_i \)'s. Since \( |S| = 6 \), the total number of appearances of elements in all \( B_i \)'s is \( 6 \times 3 = 18 \). Each \( B_i \) has two elements, so the total number of appearances is also \( 2n \). Therefore, we have:
   \[
   2n = 18 \implies n = 9
   \]

4. **Verification:**
   - Each element of \( S \) appears in exactly four of the \( A_i \)'s, and there are 6 elements in \( S \), so the total number of appearances in \( A_i \)'s is \( 4 \times 6 = 24 \).
   - Each element of \( S \) appears in exactly three of the \( B_i \)'s, and there are 6 elements in \( S \), so the total number of appearances in \( B_i \)'s is \( 3 \times 6 = 18 \).
   - Since each \( B_i \) has two elements, the total number of \( B_i \)'s is \( n = 9 \).

The final answer is \( \boxed{9} \).