Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 30 / 4 / b / Solution

The identities require $r=k$ and $k(k-1)=2\cdot 3=6$, so $k=r=3$. Take treatments $1,2,3,4$ and the four blocksEach treatment is absent from exactly one block, hence occurs three times. Each pair is absent from the two blocks missing one of its members, hence occurs together in the other two. This is the required symmetric balanced incomplete block design; here $r-\lambda =1$ is indeed a square.