Combinatorial design summation identities.

Question

Let $\mathcal{D}$ be a $(v,k,\lambda)$-design and $B\in \mathcal{D}$ an arbitary block. Let $y_i$ be the number of block which have exactly $i$ common elements with $B$, $(i < k)$. Prove that: $$\sum_{i=1}^{k-1}iy_i=k(r-1)\;\;\text{and}\;\; \sum_{i=1}^{k-1}{i\choose 2}y_i={k\choose 2}(\lambda-1)$$ I'm thinking basic combinatorial argumentation will suffice. In the first equations right-hand side since $r$ is the number of blocks containing a given point we just have to take the block $B$ out of the equation and we have $r-1$ and then multiply by $k$ which is the number of elements in each block. But I don't know how do we argue that we are calculating the same thing on the left-hand side. I have no idea why the second equation is as it is.

Answer 1

For the first one, consider the following set $$A=\{(x,C):x\in B\cap C,C\neq B\},$$ on one side, we know what every $x\in B$ is in other $r-1$ blocks so $|A|=k(r-1)$ as you have noticed. On the other hand we can split $$A=\bigcup _{i=1}^{k-1} A_i,$$ where $A_i=\{(x,C)\in A: |C\cap B|=i\},$ notice that there are $|C\cap B|\cdot y_i=i\cdot y_i$ elements in each set and they are disjoint. This is the LHS

For the second one, remember that $\lambda$ is the number of blocks in which each pair of distinct elements appear. So the right hand side is pick two elements of $B$ in $\binom{k}{2}$ and record also the sets in which these appear(except $B$ itself). Do a split like the one before and you should get your answer.