For a finite set $E$ of cardinality $n \geq 3$, let $f(n)$ denote the maximum number of $3$-element subsets of $E$, any two of them having exactly one common element. Calculate $f(n)$.
To find the maximum number of 3-element subsets of a finite set \( E \) of cardinality \( n \geq 3 \), such that any two of them have exactly one common element, we denote this maximum number by \( f(n) \).

1. **Initial Consideration**:
   Let \( abc \) be one of the 3-element subsets. If there are at least three other subsets containing \( a \), then there can be no other subsets containing \( b \) or \( c \) (since any two subsets must share exactly one common element). Hence, all other subsets must contain \( a \).

2. **Maximum Configuration**:
   This yields a maximum of:
   \[
   \left\lfloor \frac{n-3}{2} \right\rfloor + 1 = \left\lfloor \frac{n-1}{2} \right\rfloor
   \]
   This is because if we have \( k \) subsets containing \( a \), then each of the remaining \( n-1 \) elements can be paired with \( a \) in a unique subset, and we need to ensure that no two subsets share more than one element.

3. **Special Case for Small \( n \)**:
   If there are at most two other subsets containing \( a \), \( b \), or \( c \), then a maximal configuration is the symmetric Balanced Incomplete Block (BIB) design \((7,7,3,3,1)\), given by the subsets:
   \[
   \{abc, ade, afg, bdf, beg, cdg, cef\}
   \]
   This configuration is also known as the finite projective Fano plane of order 2.

4. **Verification for Small \( n \)**:
   - For \( n = 3 \), the only possible subset is \(\{abc\}\), so \( f(3) = 1 \).
   - For \( n = 4 \), the possible subsets are \(\{abc, abd\}\), so \( f(4) = 1 \).
   - For \( n = 5 \), the possible subsets are \(\{abc, abd, aec\}\), so \( f(5) = 2 \).
   - For \( n = 6 \), the possible subsets are \(\{abc, abd, aec, aed\}\), so \( f(6) = 4 \).

5. **General Case for Larger \( n \)**:
   - For \( 7 \leq n \leq 15 \), the maximum number of such subsets is given by the Fano plane configuration, so \( f(n) = 7 \).
   - For \( n \geq 15 \), the maximum number of such subsets is given by:
     \[
     f(n) = \left\lfloor \frac{n-1}{2} \right\rfloor
     \]

The final answer is \( \boxed{ f(n) = \left\lfloor \frac{n-1}{2} \right\rfloor } \) for \( n \geq 15 \), and specific values for smaller \( n \) as detailed above.