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:

\(\ds \frac {b k} v\)

\(=\)

\(\ds \frac {v - 1} {k - 1} \lambda\)

substituting for $r$

\(\ds \leadsto \ \ \)

\(\ds b k \paren {k - 1}\)

\(=\)

\(\ds \lambda v \paren {v - 1}\)

\(\ds \leadsto \ \ \)

\(\ds b \binom k 2\)

\(=\)

\(\ds \lambda \binom v 2\)

Binomial Coefficient with Two

$\blacksquare$

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