$35 \cdot 16$ A club has $x$ members, who form four committees according to the following two rules:
(1) Each member belongs to two and only two committees;
(2) Each pair of committees has one and only one member in common.

Then $x$
(A) cannot be determined.
(B) has one value and is between 8 and 16.
(C) has two values and is between 8 and 16.
(D) has one value and is between 4 and 8.
(E) has two values and is between 4 and 8.
(10th American High School Mathematics Examination, 1959)
[Solution] The number of ways to choose any two out of four committees is $C_{4}^{2}=6$. If each pair of committees appoints one member, then there are 6 people in total.
These 6 people each belong to and are only members of two committees.
Therefore, the correct choice is $(D)$.