[proofplan] We prove both identities by counting finite sets in two different ways. Let $X$ denote the finite point set and let $\mathcal B$ denote the finite collection of blocks of the given [balanced incomplete block design](/page/Balanced%20Incomplete%20Block%20Design). The first identity counts incidences between points and blocks. The second identity fixes one point $x \in X$ and counts pairs consisting of a second point and a block containing both points. The defining uniformity conditions of the design give the two expressions for each count. [/proofplan] [step:Count point-block incidences by points and by blocks] Define the incidence set \begin{align*} I := \{(x,B) \in X \times \mathcal B : x \in B\}. \end{align*} Counting $I$ by first choosing the point, each $x \in X$ contributes exactly $r$ blocks, so \begin{align*} |I| = \sum_{x \in X} r = vr. \end{align*} Counting $I$ by first choosing the block, each $B \in \mathcal B$ contributes exactly $k$ points, so \begin{align*} |I| = \sum_{B \in \mathcal B} k = bk. \end{align*} Since both expressions count the same finite set $I$, we obtain \begin{align*} vr = bk. \end{align*} [guided] The object being counted is the finite set \begin{align*} I := \{(x,B) \in X \times \mathcal B : x \in B\}, \end{align*} whose elements are incidences between a point and a block. We count this set in two ways. First fix a point $x \in X$. By the definition of the parameter $r$, the point $x$ lies in exactly $r$ blocks. Therefore exactly $r$ ordered pairs in $I$ have first coordinate $x$. Since $X$ has $v$ elements, summing over all points gives \begin{align*} |I| = \sum_{x \in X} r = vr. \end{align*} Second fix a block $B \in \mathcal B$. By the definition of the parameter $k$, the block $B$ contains exactly $k$ points. Therefore exactly $k$ ordered pairs in $I$ have second coordinate $B$. Since $\mathcal B$ has $b$ elements, summing over all blocks gives \begin{align*} |I| = \sum_{B \in \mathcal B} k = bk. \end{align*} Both computations count the same finite set $I$, so their results are equal: \begin{align*} vr = bk. \end{align*} [/guided] [/step] [step:Count blocks through a fixed point and one other point] Since the point set $X$ is nonempty by hypothesis, fix a point $x \in X$. Define the finite set \begin{align*} J_x := \{(y,B) \in (X \setminus \{x\}) \times \mathcal B : x \in B \text{ and } y \in B\}. \end{align*} Counting $J_x$ by first choosing a block containing $x$, there are exactly $r$ such blocks, and each contains exactly $k-1$ points of $X \setminus \{x\}$. Hence \begin{align*} |J_x| = r(k-1). \end{align*} Counting $J_x$ by first choosing $y \in X \setminus \{x\}$, there are exactly $v-1$ choices of $y$, and each pair $\{x,y\}$ lies in exactly $\lambda$ blocks. Hence \begin{align*} |J_x| = \lambda(v-1). \end{align*} Equating the two counts of $J_x$ gives \begin{align*} r(k-1)=\lambda(v-1). \end{align*} [/step]