Family of subsets - cardinality 2 Let $n\in\mathbb N, k\in\{1,\dots,n\}$ and $C = \lbrace A_1, \ldots ,A_r \rbrace$, where $A_i\subset \lbrace 1, 2, \ldots n \rbrace$, such that:
$$\forall i \neq j \ : \ |A_i \cap A_j|=k$$
Show that $|C| \leq n$.
 A: Missing from the statement is still $A_i \neq A_j$, but that's probably obvious...  Also implicitly here we have $| C | = r$.
The solution follows a similar line as the Oddtown problem.  Let the fixed value be $l$.  For each subset $A_i$ make an $n$ dimensional column vector $V_i$, with $1$ in the $i$th entry if it contains the $i$th element, $0$ otherwise.  Then the $r\times n$ matrix $M$ composed of the transposes of the column vectors is a linear transformation from $\mathbb{R}^n$ to $\mathbb{R}^r$.  Applying $M$ to any $V_i$, we get a vector with $l$ in each entry except for the $i$th entry, which has some value $m_i > l$ (since otherwise we'd have $A_i = A_j \forall j$).  So now we just need to show that it spans a space of dimension $r$, forcing $r \leq n$.
Then arrange the results into an $r \times r$ matrix.  This can be expressed as $D + lJ$ where $D$ is diagonal with entries $m_i - l$, and $J$ is the $r \times r$ matrix with all entries $1$.  Then the determinant is $\det(D)\det(I+c_m r_m)$ where $c_m$ is a column vector with $1/(m_i - l)$ and $r_m$ is the row matrix with entries $l$.  Then by Sylvester's determinant theorem, this is just $\det(D)(1 + r_m c_m)$, which is non-zero since both the $r_m$ and $c_m$ entries are positive.
