I could not solve an exercise in the book of Stinson. Here is the question:
Let B0 be a block in a $(v, k, 1)$-BIBD, say $(X, B)$.
$(a)$ Find a formula for the number of blocks B ∈ $B$ such that |B ∩
B0| = 1.
$(b)$ Use your formula to show that $b ≥ k(r −1)+1$ if a $(v, k, 1)$-BIBD
exists.
$(c)$ Using the facts that $vr = bk$ and $v = r(k − 1) + 1$, deduce that
$(r−k)(r−1)(k−1) ≥ 1$, and hence $r ≥ k$,which implies Fisher’s
Inequality.
For part $(a)$,
In symmetric case(b=v): I know that the number of blocks B ∈ B such that |B ∩ B0| = 1 is equal to $b-1$ since multiplication of a block B0 with different blocks will give $\lambda=1$.
But, for $b\neq v$, I do not know the formula.
Thanks for any help
For every $x\in X$, let $\mathcal{B}_x:=\{B\in\mathcal{B}\;|\;x\in B\}$ and ${\mathcal{B}_x}':=\{B\setminus\{x\}\;|\;B\in\mathcal{B}\}$. Then $X\setminus\{x\}$ is the disjoint union of all elements of ${\mathcal{B}_x}'$, i.e. $$v-1=(k-1)\cdot|{\mathcal{B}_x}'|=(k-1)\cdot|\mathcal{B}_x|.$$ Therefore, for a fixed $B_0\in\mathcal{B}$, we have $$\{B\in\mathcal{B}\;|\;|B\cap B_0|=1\}=\bigcup_{x\in B_0}(\mathcal{B}_x\setminus\{B_0\}),$$ and the number in question is $$\begin{align} |\,\{B\in\mathcal{B}\;|\;|B\cap B_0|=1\}\,| &=|\,\bigcup_{x\in B_0}(\mathcal{B}_x\setminus\{B_0\})\,|\\ &=\sum_{x\in B_0}(\,|\mathcal{B}_x|-1\,)\\ &=k\;\cdot\;(\,\frac{v-1}{k-1}-1\,)\\ &=k\;\cdot\;(\,r-1\,)\\ &=\frac{vk-k^2}{k-1} \end{align}$$