# Necessary Condition for Existence of BIBD

## Theorem

Let there exist be a BIBD with parameters $v, b, r, k, \lambda$.

Then the following are true:

- $(1): \quad b k = r v$

- $(2): \quad \lambda \left({v-1}\right) = r \left({k-1}\right)$

- $(3): \quad \displaystyle b \binom k 2 = \lambda \binom v 2$

- $(4): \quad k < v$

- $(5): \quad r > \lambda$

All of $v, b, r, k, \lambda$ are integers.

Some sources prefer to report $(3)$ as:

- $\displaystyle b = \frac { {v \choose 2} }{ {k \choose 2} }\lambda$

which is less appealing visually, and typographically horrendous.

## Proof

- $(1)$: We have by definition that each point is in exactly $r$ blocks, and each block is of size $k$.

We have that $b k$ is the number of blocks times the size of each block.

We also have that $r v$ is the number of points times the number of blocks each point is in.

The two must clearly be equal.

- $(2)$: Comparing the left and right hand sides of the equation we can see that:

LHS: An arbitrary point must be paired with $v-1$ other points.

If $\lambda>1$ then every point is paired $\lambda \left({v-1}\right)$ times.

RHS: An arbitrary point is paired with $k-1$ other points for each of the $r$ blocks it is in.

Therefore it is paired $r \left({k-1}\right)$ times.

Both values give the number of times an arbitrary point is paired, therefore LHS = RHS.

- $(3)$: From equation $(1)$, we have that $r = \dfrac {bk} v$

From $(2)$ we have that $r = \dfrac {v-1} {k-1} \lambda$.

So:

\(\displaystyle \frac {bk} v\) | \(=\) | \(\displaystyle \frac {v-1} {k-1} \lambda\) | $\quad$ substituting for $r$ | $\quad$ | |||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle b k \left({k-1}\right)\) | \(=\) | \(\displaystyle \lambda v \left({v-1}\right)\) | $\quad$ | $\quad$ | ||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle b \binom k 2\) | \(=\) | \(\displaystyle \lambda \binom v 2\) | $\quad$ from Binomial Coefficient with Two | $\quad$ |

$\blacksquare$