Necessary Condition for Existence of BIBD

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Theorem

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

Then the following are true:

$(1): \quad b k = r v$
$(2): \quad \lambda \paren {v - 1} = r \paren {k - 1}$
$(3): \quad b \dbinom k 2 = \lambda \dbinom v 2$
$(4): \quad k < v$
$(5): \quad r > \lambda$

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


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

$b = \dfrac {\dbinom v 2} {\dbinom k 2} \lambda$

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:

left hand side: An arbitrary point must be paired with $v-1$ other points.

If $\lambda>1$ then every point is paired $\lambda \paren {v - 1}$ times.


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

Therefore it is paired $r \paren {k - 1}$ times.

Both values give the number of times an arbitrary point is paired, therefore left hand side = right hand side.


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

From $(2)$ we have that:

$r = \dfrac {v - 1} {k - 1} \lambda$

So:

\(\displaystyle \frac {b k} v\) \(=\) \(\displaystyle \frac {v - 1} {k - 1} \lambda\) substituting for $r$
\(\displaystyle \leadsto \ \ \) \(\displaystyle b k \paren {k - 1}\) \(=\) \(\displaystyle \lambda v \paren {v - 1}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle b \binom k 2\) \(=\) \(\displaystyle \lambda \binom v 2\) Binomial Coefficient with Two

$\blacksquare$