Symmetrical combinatoral block design Suppose that $(V,\mathcal{B}$) is a symmetric block design of type $2-(v,k,\lambda)$. Pick an arbitrary block $B \in \mathcal{B}$. Let $B_1,...,B_{v-1}$ be a list of the remaining blocks and let $n_1,...,n_{v-1}$ be respectively the number of elements that $B$ has in common with these blocks. Show that $$n_1+...+n_{v-1} = (v-1)\lambda$$


I'm quite confused as to where to start this problem. I know that for a symmetric block design, $b=v$, so we know that the sum is equal to $(b-1)\lambda$, which essentially means that the each intersection has $\lambda$ elements. I'm unsure as to why that's true, or how to go about proving it. Any hints?
So we have $v$ blocks (in the set $V$ with $v$ elements), each having $k$ elements and each element in $V$ apears in $k$ blocks and every pair of elements $\{x,y\}$ appears in exactly $\lambda $ blocks.

So if we make a double counting between the blocks $B_1,...B_{v-1}$ and the elements in $B$ we have $$n_1+...n_{v-1} = (k-1)k$$ 
since each of $k$ elements in $B$ appears exactly $k$ times (once in $B$ and the rest in $k-1$ sets).

But on the other side each pair $\{x,y\}$ appears in exatly $\lambda $ blocks, so we have again by double counting between the set of all pairs of elements in $V$ and the set $\mathcal{B}$: $${v\choose 2}\cdot \lambda = v\cdot {k\choose 2}\implies (v-1)\cdot \lambda = k(k-1)$$

and thus the conclusion.
I would interpret the left hand side of the equation as counting all pairs $(B_i, p)$ such that $p\in B_i$ and $p\in B$. The left hand side of the equation does this by considering all blocks $B_i$, besides $B$, and counting their intersections with $B$. Another way to count these pairs is to instead consider all points $p\in B$, and then count how many blocks besides $B$ contain $p$. If you do that, and recall the few equalities associated with block designs and symmetric block designs in particular, you should get your answer.