In a BIBD, why is it the case that $r>\lambda$?

Question

Given a BIBD $(v,b,r,k,\lambda)$, I'm trying understand the proof of Fisher's inequality, but I'm stuck at the step where the assertion $r>\lambda$ is made. But why? The only pieces of information I have is $vr=bk$ and $\lambda(v-1)=r(k-1)$.

Answer 1

Every point is in precisely $r$ blocks.

Every pair of distinct points is in precisely $\lambda$ blocks.

Since $\{p,q\}\subset S\implies p\in S$, it is unavoidable that $r\ge\lambda$.

Now suppose that $r=\lambda$. Then for any point $p$, occurring in $r$ blocks, every pair of points $\{p,q\}$ must appear in all $r$ blocks; so every other point uses the same $r$ blocks, and $b=r$ and $v=k$, making all blocks complete and the design degenerate.