Balanced Incomplete Block Design (BIBD) 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?
 A: 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. 
A: The first sentence:
"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 $1$" 
sounds a little bit confusing.
Using Montgomery's notation, the relations $ar=bk$ and $r(k-1)=\lambda(a-1)$ define a BIBD. 
That is, given a generic treatment $A$, there are $r$ blocks containing it and $k-1$ other treatments in each of these blocks, so there are $r(k-1)$ pairs.
We can also say that treatment $A$ can be paired with $a-1$ treatments and each of this pairs can appear $\lambda$ times.
