Prove that $\lambda(v-1) = r(k-1)$

Question

This is to do with balanced incomplete block design. Some homework exercise wants me to prove the relation $$\lambda(v-1) = r(k-1)$$

$v$ is the number of elements in your ground set.

$r$ is the number of blocks containing a given element

$k$ is the number of elements in each block.

$\lambda$ is the number of blocks containing each pair of elements.

I don't really know where to begin with this, especially since this is the only equation that I have ever used $\lambda$ for.

Answer 1

I assume here (although you should really specify) the following meanings for the parameters:

• $v$ is the number of elements in your ground set.

• $r$ is the number of blocks containing a given element

• $k$ is the number of elements in each block.

• $\lambda$ is the number of blocks containing each pair of elements.

Assuming that this is correct: notice that if $x$ is a fixed element of your ground set, then $\lambda(v-1)$ is the number of ways of choosing a pair $(y,B)$, where $y\neq x$ and $B$ is a block such that $x,y\in B$.

Can you prove that the number of such pairs is also $r(k-1)$?

Answer 2

I assume you are working on a two design, if that is so. Then you can use the gen eqn: $(v-i)\lambda_{i+1} = (k-i)\lambda_i$, $i$ is between $0$ and $t-1$. (Boggs and White p.57) There you'll have $i=1$, $(v-1)\lambda_2 = (k-1)\lambda_1$, and we know that $\lambda_2 = \lambda_t = \lambda$ and $\lambda_1 = r$. Thus you have your equation :)