Discussion
Identify the treatments \(x_{1}, x_{2}, x_{3}\) and \(x_{4}\) from blocks 1, 2, 3, 4 respectively so that the design is BIBD :
Block 1 : A, B, C, \(x_{1}\)
Block 2 : A, \(x_{2}\), C, E
Block 3 : A, B, D, \(x_{3}\)
Block 4 : A, \(x_{4}\), D, E
Block 5 : B, C, D, E
Answer: B
The design is a BIBD with \(x_{1}\) = D, \(x_{2}\) = B, \(x_{3}\) = E, \(x_{4}\) = C.
With these values the five blocks are {A, B, C, D}, {A, B, C, E}, {A, B, D, E}, {A, C, D, E} and {B, C, D, E}: every 4-element subset of the five treatments, each exactly once. So v = b = 5, k = 4, every treatment appears in r = 4 blocks, and every pair of treatments appears together in \(\lambda = 3\) blocks. Check: \(\lambda(v - 1) = r(k - 1)\) gives \(3 \times 4 = 4 \times 3\).
Each of the other choices repeats a treatment within a block, which a BIBD does not allow: the first puts B twice in block 1, the third puts C twice in block 1, and the last puts C twice in block 2 and D twice in block 3.
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