Number of club members if they are organized into four committees if each member is in exactly two committees and any two committees share one member A club with $x$ members is organized into four committees such that
$(a)$ each member is in exactly two committees,
$(b)$ any two committees have exactly one member in common.
Then $x$ has
$(A)$ exactly two values both between $4$ and $8$
$(B)$ exactly one value and this lies between $4$ and $8$
$(C)$ exactly two values both between $8$ and $16$
$(D)$ exactly one value and this lies between $8$ and $16$.
My thought is option $B$ and $x = 6$.  Please somebody tell me am I right or wrong?
 A: Claim: $x=6$
Let $S:=[x]\left(=\{1,2,\dots ,x\}\right)$ where $x$ is some fixed natural number. Let $C_i\subset S$ be the $i$ th commitee.Let $f$ be a function from the set of unordered pair of distinct commitees to members of the club defined as,
$$f\left( (C_i,C_j)\right)=C_i\cap C_j$$
$f$ is injective by (a). $f$ is surjective by (b).
Hence, $|S|=\binom{4}{2}=6$. 

Here is a construction:


*

*Let $C_1\cap C_2=1$ wlog. Then, $1\notin C_i \quad \forall \, i\in
   [4]\setminus \{1,2\}$.

*Let $C_2\cap C_3=2$ wlog. Then, $2\notin C_i \quad \forall \, i\in
   [4]\setminus \{2,3\}$.

*Let $C_3\cap C_4=3$ wlog. Then, $3\notin C_i \quad \forall \, i\in
   [4]\setminus \{3,4\}$.

*Let $C_1\cap C_3=4$ wlog. Then, $4\notin C_i \quad \forall \, i\in
   [4]\setminus \{1,3\}$.

*Let $C_1\cap C_4=5$ wlog. Then, $5\notin C_i \quad \forall \, i\in
   [4]\setminus \{1,4\}$.

*Let $C_2\cap C_4=6$ wlog. Then, $6\notin C_i \quad \forall \, i\in
   [4]\setminus \{2,4\}$.
A: Let the committees be $A_1, A_2, A_3, A_4$ then $|A_1\cup A_2 \cup A_3 \cup A_4|=n$. There are $\binom{4}{2}=6$ pairwise intersections each of size $1$ and each intersection of triples must be empty since if not some member would belong to $3$ committees, a contradiction. If we make a $4 \times n$ table where rows are committees and columns are members and mark a $1$ whenever member $x_j$ belongs to committee $A_i$ we notice that the $i^{th}$ row sum is $|A_i|$ and the $j^{th}$ column sum is $2$ (since each member belongs to exactly $2$ committees) so equating row and column sums we get $\sum_{i=1}^{4}|A_i|= 2n.$
Putting this information together and using the inclusion-exclusion formula we get $n =6$
