If treatment $1$ appears in $r$ blocks and there are $(k-1)$ other treatments in each of those blocks, there are $r(k-1)$ observations in a block containing treatment $1$.
But the following Table does not follow the above statement but the table is said to be BIBD:
$$
Block
$$
$$
\begin{array}{l}
1 & 2 &3 &4 \\
\hline
A & A &A & B \\
B & B &C & C \\
C & D &D & D \\
\end{array}
$$
here, $k=3$ , $r=3$
But each block doesn't contain $r(k-1)=3(3-1)=6$ observations rather each block is containing $3$ observations.
Where am i doing the mistake?
Best Answer
In a balanced incomplete block design (BIBD): all the block have the same size $k$, all pairs of treatments occur together in the same block the same number of times $\lambda$ and there are $\nu$ treatments. The number of replicates of each treatment is $r$.
In your example we have: $$ k=3 \\ \lambda= 2 \qquad \text{example: $A,B$ together in blocks 1,2} \\ \nu = 4 \qquad A,B,C,D \\ r=3 $$ Then $n$, the number of experimental runs, is given by $n= \nu \cdot r = 4\cdot 3 = 12$. It can also be calculated as $\binom{\nu}{2}\cdot \lambda = 6 \cdot 2 = 12$. Now you can use these relations to check what you said.