Theorem $\mathbf{3.4.}$ In a block design of type $2-(v,k,\lambda)$ every element lies in precisely $r$ blocks, where $$r(k-1)=\lambda(v-1)\textit{ and }bk=vr\;.$$ The letter $r$ stands for “replication number”.
Proof: We proceed by double counting. Fix an arbitrary $v_0\in V$. This time we count the number of pairs $(T,B)$ with $T$ being a $2$-element subset of $V$ containing $v_0$, and $T\subseteq B\in\mathcal{B}$. This number is either $\lambda(v-1)$, by starting with the possible choices for $T$, or $r(k-1)$, considering the choices for $B$ first (there are $r$ blocks that contain $v_0$, and there are $k-1$ other elements in a given block which can be used to form the pair $T$). This yields the first equality. The second follows from considering the number of pairs $(T,B)$ with $T$ being a $1$-element subset of $B$. It is either $bk$ since each of the $b$ blocks contains $k$ elements, or $vr$ since each of the $v$ elements of $V$ occurs in $r$ blocks of $\mathcal{B}$. This concludes the proof of the second part of the statement. $\quad\square$
I am trying to make sense of the above theorem and its proof. While the proof itself is straightforward, I am having trouble what $(T,B)$ represents, I can't quite get my head around what I am counting (I have this problem with a lot of double counting proofs, so I think if I can make thorough sense of this, I should be on my way).
Thanks for your help.
Since $v$ is being used to denote the cardinality of $V$, I’m going to use $x$ and $x_0$ for elements of $V$.
Let $\mathcal{T}$ be the collection of all $2$-element subsets of $V$, and let $x_0$ be some fixed element of $V$. Then we’re counting
$$\mathcal{P}=\{\langle T,B\rangle\in\mathcal{T}\times\mathcal{B}:x_0\in T\subseteq B\}\;,$$
the set of all ordered pairs whose first component is a $2$-element subset $T$ of $V$ containing $x_0$, and whose second component is a block $B$ of the design that has $T$ as a subset. Why this rather strange collection $\mathcal{P}$? Basically because it works: we can count it in two ways to get the desired equality.
On the one hand we know that there are $v-1$ members of $\mathcal{T}$ that contain $x_0$, since we can pair $x_0$ with any of the other $v-1$ members of $V$. We further know that each of these $v-1$ pairs is a subset of $\lambda$ different blocks $B$, so for each $x\in V\setminus\{x_0\}$, there are $\lambda$ different blocks $B$ such that the pair $\langle\{x_0,x\},B\rangle\in\mathcal{P}$. Thus, there are altogether $(v-1)\cdot\lambda=\lambda(v-1)$ ordered pairs in $\mathcal{P}$.
On the other hand we know that $\mathcal{B}$ has $r$ blocks that contain $x_0$. If $B$ is one of those blocks, then $|B|=k$, so $B$ has $k-1$ elements other than $x_0$. If $x$ is any one of these $k-1$ elements, then $\langle\{x_0,x\},B\rangle$ is one of the elements of $\mathcal{P}$, and every element of $\mathcal{P}$ can be obtained in this way, so $\mathcal{P}$ must contain $r(k-1)$ ordered pairs.
Putting the two together, we see that $\lambda(v-1)=r(k-1)$.
To get the second equation we actually count a different set. Let
$$\mathcal{P}'=\{\langle x,B\rangle\in V\times\mathcal{B}:x\in B\}\;,$$
the let of all ordered pairs whose first component is an element $x$ of $V$, and whose second element is a block containing $x$.
On the one hand there are $v$ possible choices of $x\in V$ for the first component of a member of $\mathcal{P}'$, and each of them is by definition in exactly $r$ blocks, so there are $vr$ ordered pairs in $\mathcal{P}'$.
On the other hand there are $b$ blocks, and each has $k$ elements, so there are $bk$ ordered pairs in $\mathcal{P}'$.
Putting the two together, we find that $bk=vr$.
You could think about arguments in terms of matrices. I’ll start with the second one, since it’s a little simpler. Imagine a $v\times b$ matrix $M'$ with a row for each point of $V$ and a column for each block. If element $x$ is in block $B$, put a $1$ in row $x$, column $B$ of $M'$, and otherwise put a $0$ there. Now count the $1$s in $M'$.
Each element of $V$ is in $r$ blocks, so there are $r$ $1$s in each row of $M'$. And there are $v$ rows, so $M'$ has $vr$ $1$s altogether.
Alternatively, each block contains $k$ points of $V$, so there are $k$ $1$s in each column of $M'$. And $M'$ has $b$ rows, so it has $bk$ $1$s altogether.
Thus, $vr=bk$.
For the first equation we construct a $(v-1)\times r$ matrix $M$ with a row for each point of $V\setminus\{x_0\}$ and a column for each block containing $x_0$. If $x\in V\setminus\{x_0\}$ and $B$ is a block containing $x_0$, we put a $1$ in row $x$, column $B$ if $\{x_0,x\}\subseteq B$, and otherwise we put a $0$ there. Again we count the $1$s.
Each pair of points of $V$ is contained in $\lambda$ blocks, so there are $\lambda$ $1$s in each row of $M$. $M$ has $v-1$ rows, so it must have $\lambda(v-1)$ $1$s.
Each block containing $x_0$ contains $k-1$ other points of $V$, so there are $k-1$ $1$s in each column of $M$. $M$ has $r$ columns, so it must have $r(k-1)$ $1$s.
Thus, $\lambda(v-1)=r(k-1)$.