Definition 1 (see [14].)A t-(v, k, λ) design, or a t-design is a pair $$, where $$ is a v-set of points and $$ is a collection of k-subsets (blocks) of $$ with the property that every t-subset of $$ is contained in exactly λ blocks. The parameter λ is called the index of the design.

A necessary condition for the existence of a t-(v, k, λ) design is known [13] to be the following equation:

The problem of determining the values of k, t, and $$ for which this condition is also sufficient is not yet solved completely, and less is known about configurations with t = 3.

Example 1. A 3-(8,4,1) design on the set $$ = {1,2, …, 8} is given by the blocks: {1,2,5,6}, {3,4,7,8}, {1,3,5,7}, {2,4,6,8}, {1,4,5,8}, {2,3,6,7}, {1,2,3,4}, {5,6,7,8}, {1,2,7,8}, {3,4,5,6}, {1,3,6,8}, {2,4,5,7}, {1,4,6,7} and {2,3,5,8}.

The concepts of a GDD and a 3-design can be merged to define a 3-GDD as follows.

Definition 2. A 3^′-GDD (n, m, k, Λ₁, Λ₂), for m > 2, is a set $$ of mn symbols partitioned into m parts of size n called groups together with a collection of k-subsets of $$ called blocks, such that

• (i)

  Every 3-subset of each group occurs in Λ₁ blocks

• (ii)

  Every mixed 3-subset, meaning either two symbols are from one group and one symbol from the other group or all three symbols are from different groups, occurs in Λ₂ blocks

The above mentioned definition was given in [14, 15], in which case a 3-subset of $$ has only two choices, all symbols from one of the two groups or one symbol from a group and two symbols from the other group. Such GDD was denoted by a 3-GDD (n, 2, 4, λ₁, λ₂).

• (i)

  The necessary conditions are sufficient for the existence of a 3-GDD (n, 2, 4, λ₁, λ₂) for n ≡ 1,7,9(mod 12) [14]

• (ii)

  Except possibly when n ≡ 1,3(mod 6), n ≠ 3, 7, 13, the necessary conditions are sufficient for the existence of a 3-GDD (n, 2, 4, λ₁, λ₂) and λ₁ > λ₂ [15]

Lemma 3 (see [14].)If a 3-(2n, 4, λ₂) and a 3-(2n, 4, λ₁ − λ₂) exists, then a 3-GDD (n, 2, 4, λ₁, λ₂) exists.

Example 2. Here, $$ = {1,2,3, a, b, c}, G₁ = {1,2,3} and G₂ = {a, b, c}, the blocks of a 3-GDD (3,2,4,3,1) are: {1,2,3, a}, {1,2,3, b}, {1,2,3, c}, {a, b, c, 1}, {a, b, c, 2} and {a, b, c, 3}.

Following necessary conditions (Table 1, where the values of λ₁ and λ₂ are given modulo 6) and the existence results of a 3-GDD (n, 2,4, λ₁, λ₂) are given in [14, 15].

Definition 4. A 3-PBIBD (partially balanced incomplete block design), 3-PBIBD (n, m, k, θ₁, θ₂, θ₃) is a collection of k-subsets of an mn set $$, where $$ is partitioned in to m groups of order n such that

• (i)

  Every triple formed from symbols of only a single group occurs in θ₁ blocks

• (ii)

  Every triple formed from symbols of only two groups occurs in θ₂ blocks

• (iii)

  Every triple formed from symbols of all three groups occurs in θ₃ blocks

Definition 5. A 3-GDD (n, m, k, μ₁, μ₂) is a pair ($$, $$), where $$ is a set of mn elements partitioned into mn-subsets (groups) and $$ is a collection of k-subsets (blocks) of $$ such that

• (i)

  Every triple occurs in exactly μ₁ blocks if it contains elements from at most 2 groups

• (ii)

  It occurs in exactly μ₂ blocks if it has all three elements from different groups

Definition 5 is the subject matter of this paper and we explicitly consider the case in which m = 3 and k = 5. Such GDD is denoted by 3-GDD (n, 3, 5, μ₁, μ₂).

Remark 6. A 3-GDD (n, 3, 5, μ₁, μ₂) is a 3-PBIBD (n, 3, 5, θ₁, θ₂, θ₃), where θ₁ = θ₂, denoted by μ₁, while θ₃ is denoted by μ₂.

When μ₁ = μ₂, a 3-GDD (n, 3, 5, μ₁, μ₂) is actually a 3-(3n, 5, μ₁). It follows that a 3-GDD (n, 3, 5, μ₁, μ₂) exists if and only if a 3-(3n, 5, μ₁) exists.

The next three sections of this paper discuss our research findings.

Theorem 7. Given a 3-GDD (n, 3,5, μ₁, μ₂),

Proof. The abovementioned results can be proved as follows:

• (1)

  We count the number of triples containing a fixed element x in the design in two ways.

•  

  First, given an element x, it appears in $$ triples of the type (3, 0) and in 2n(n − 1) + 2 $$ triples of the type (2,1) (there are 2n(n − 1) triples of the type (2,1) where x and an element from the same group as x appear with an element of another group, and there are 2 $$ triples of type (2,1) where x appears with two elements from another group). So, in sum, x is in $$μ₁ triples of the form (3,0) or (2,1) in the design.

•  

  Similarly, there are n² triples of the type (1,1,1) containing x, and are repeated μ₂ times.

•  

  All together, there are $$μ₁ + n²μ₂ triples containing x.

•  

  Second, in every block containing x, there are 6 triples containing x and x occurs in r blocks, which means there are 6r triples containing x in the design.

•  

  Hence, 6r = $$μ₁ + n²μ₂, and r = ((n − 1)(7n − 2)μ₁ + 2n²μ₂)/12

• (2)

  Again, counting in two ways, in a design with block size 5 and b blocks, there are $$ = 10b triples. On the other hand, there must be exactly $$ triples of type (3,0), 3n²(n − 1)μ₁ triples of type (2,1), and n³μ₂ triples of type (1,1,1) in the design.

•  

  Therefore, 10b = $$ + 3n²(n − 1)μ₁ + n³μ₂ gives equation (3).

• (3)

  Let (x₁, x₂) be a first associate pair and occurs in λ₁ blocks. The triples containing (x₁, x₂) can only be of type (3,0) or (2, 1). There are (n − 2) triples of the type (3,0) and 2n triples of the type (2,1) containing (x₁, x₂). Now, since these triples occur μ₁ times and since the block containing a first associate pair, say (x₁, x₂) contains three triples containing (x₁, x₂), we have: λ₁ = ((3n − 2)/3)μ₁.

• (4)

  Let (x, y) be a second associate pair. This pair occurs in triples of the type (2,1) and (1, 1, 1) only. It occurs in 2(n − 1)μ₁ triples of type (2,1) and nμ₂ triples of type (1, 1, 1).

Each of the λ₂ blocks containing the pair (x, y) has three triples containing (x, y).

Hence, 3λ₂ = 2(n − 1)μ₁ + nμ₂ and λ₂ = (2(n − 1)μ₁ + nμ₂)/3.

Corollary 8. For μ₁ ≡ 0(mod 3), a 3-GDD (n, 3, 5, μ₁, μ₂) does not exist.

Proof. From equation (3) in Theorem 7, λ₁ is not an integer when μ₁ ≡ 0(mod 3).

Corollary 9. In a 3-GDD (n, 3, 5, μ₁, μ₂), λ₁ ≠ 0.

Proof. As the number of groups is less than the block size, μ₁ ≠ 0. Again from (3) in Theorem 7, λ₁ is a multiple of μ₁, and hence λ₁ ≠ 0

From the original necessary conditions in Theorem 7, when μ₁ = μ₂ = λ, we get 3-(3n, 5, λ), and from (4) and (5), we must have λ ≡ 0(mod 3). In addition, from (2) and (3) in Theorem 7, the following result follows.

Corollary 10. The necessary conditions for the existence of 3-(3n, 5, λ) are satisfied only under the following cases:

• (i)

  If n ≡ 2, 7, 10, 14, 15(mod 20), then λ ≡ 0(mod 3)

• (ii)

  If n ≡ 0, 4, 5, 9, 12, 17(mod 20), then λ ≡ 0(mod 6)

• (iii)

  If n ≡ 3, 6, 11, 18(mod 20), then λ ≡ 0(mod 15)

• (iv)

  If n ≡ 1, 8, 13, 16(mod 20), then λ ≡ 0(mod 30)

Theorem 11. Given a 3-GDD (n, 3, 5, μ₁, μ₂),

• (i)

  μ₂ < 3μ₁

• (ii)

  b ≥ (3n²(n − 1)μ₁ + n³μ₂)/15

Proof.

• (i)

  In a 3-GDD (n, 3, 5, μ₁, μ₂), triples of the type (1,1,1) occur in the blocks of type (3,1,1) and (2,2,1) only. In each of these blocks which can have a (1,1,1) triples, the number of (2,1) triples are more than the number of (1,1,1) triples.

•  

  Hence, 3n²(n − 1)μ₁ > n³μ₂, implies μ₂ < 3(n − 1)μ₂/n < 3μ₁.

• (ii)

  We calculated the value of b = (n/20)((n − 1)(7n − 2)μ₁ + 2n²μ₂) in Theorem 7, if we do not include the triples of the form (3, 0), in the calculations, we obtain the following equation:

  $$ (6)

Theorem 12. For n ≤ 5, a 3-GDD (n, 3, 5, μ₁, 0) does not exist.

Proof. As μ₂ = 0, blocks are of configuration (5,0), (4,1) and (3,2) only. The maximum number of (2,1) triples occur in (3,2) configuration blocks and ratio is 9 : 1 (in each block of configuration (3,2), there are nine (2,1) and one (3,0) triples). Counting the number of (2,1) triples (3n²(n − 1)μ₁) and the number of (3,0) triples (n(n − 1)(n − 2)/2), their ratio must be less than or equal to 9.

Thus, 3n²(n − 1)μ₁/n(n − 1)(n − 2)/2 ≤ 9⇒6n/n − 2 ≤ 9, which gives n ≥ 6.

From (4) and (5) in Theorem 7, the necessary conditions are satisfied under the following conditions.

Lemma 13. In regards to

• (i)

  λ₁, for every n, μ₁ ≡ 0(mod 3) and μ₂ is free

• (ii)

  λ₂, when

  • (i)

     n ≡ 0(mod 3), μ₁ ≡ 0(mod 3) and μ₂ is free

  • (ii)

     n ≡ 1(mod 3), μ₁ is and μ₂ ≡ 0(mod 3)

  • (iii)

    n ≡ 2(mod 3), μ₁ + μ₂ ≡ 0(mod 3)

As the values of b and r must also be integers, Table 2 is a table of congruence restrictions for a 3-GDDs with 3 groups and block size 5 (all values are considered to be in terms of (mod 60) unless otherwise stated).

where

• (i)

  ⋏₁ = 4μ₁ + μ₂ ≡ 0(mod 5), ⋏₂ = μ₁ + μ₂ ≡ 0(mod 10), ⋏₃ = 9μ₁ + μ₂ ≡ 0(mod 10), ⋏₄ = 4μ₁ + μ₂ ≡ 0(mod 10), ⋏₅ = 5μ₁ + μ₂ ≡ 0(mod 10), ⋏₆ = 8μ₁ + 3μ₂ ≡ 0(mod 10), ⋏₇ = μ₁ + μ₂ ≡ 0(mod 2), and ⋏₈ = μ₁ + μ₂ ≡ 0(mod 5)

• (ii)

  ∗₁ = 5μ₁ + 3μ₂ ≡ 0(mod 6), ∗₂ = 3μ₁ + 2μ₂ ≡ 0(mod 6), ∗₃ = 3μ₁ + μ₂ ≡ 0(mod 6), and ∗₄ = 2μ₁ + 3μ₂ ≡ 0(mod 6).

Lemma 14. For

• (i)

  n ≡ 0(mod 10), μ₁ and μ₂ are free in regards to the number of blocks b

• (ii)

  n ≡ 0(mod 3) and μ₁ ≡ 0(mod 3), λ₁&λ₂ are integers for any chosen μ₂

• (iii)

  n ≡ 1,2(mod 3), μ₁ ≡ 0(mod 3) and μ₂ ≡ 0(mod 3), λ₁ & λ₂ are integers

• (iv)

  n ≡ 1,5(mod 12) and μ₂ ≡ 0(mod 6) or n ≡ 2,10(mod 12) and μ₂ ≡ 0(mod 3), r is an integer for any chosen μ₁

• (v)

  n ≡ 1,2,4,7,10(mod 12) and n ≡ 5,17,41,53(mod 60), μ₂ is multiple of 3

• (vi)

  n ≡ 6(mod 10) and n ≡ 1(mod 20), μ₂ is multiple of 5

Theorem 15.

• (a)

  All blocks of a design are of type (3,2) constitute a 3-PBIBD (n, 3, 5, n(n − 1), 3(n − 1)(n − 2)/2, 0), where the number of blocks of such design is $$$$

• (b)

  There exists a 3-PBIBD (n, 3, 5, (n − 3)(n − 4)/2, 0, 0)

Proof.

• (a)

  When all blocks are of type (3,2), the number of blocks of such design is $$$$, where $$ gives the number of 3-subsets of 3 groups and $$ gives the number of 2-subsets for each choice of a 3-subset. This also tells the number of blocks containing every (3,0) triple is $$ = n(n − 1).

•  

  Similarly, let {a, b, x} be (2,1) triple, where the pair {a, b} is from say G₁ and the element x is from G₂. There are (n − 2)(n − 1) triples containing {a, b, x} (one element each from G₁ and G₂) or there are $$ triples containing {a, b, x} (3-subsets of G₂), and hence in total there are 3(n − 1)(n − 2)/2 blocks of size five containing {a, b, x}.

• (b)

  When all blocks are of type (5,0), the number of blocks containing every (3,0) triple is $$ = (n − 3)(n − 4)/2.

Example 3. A 3-GDD (6, 3, 5, 30, 0) exists.

In Theorem 15 (a), if n = 6, then, θ₁ = θ₂ = n(n − 1) = 3(n − 1)(n − 2)/2 = 30.

Blocks are of type (4, 1) if there are four elements from a single group and one element from other group.

Theorem 16. If all the blocks are of type (4,1), then a 3-GDD (n, 3, 5, μ₁, 0) does not exist.

Proof. when all blocks are of type (4,1), a fixed (3,0) triple occurs in 2n(n − 3) and a fixed (2,1) triple occurs in $$ blocks.

Hence, 2n(n − 3) = $$, which is true only for 3n = −2, which is impossible.

In Theorem 12, a 3-GDD (n, 3, 5, μ₁, 0) does not exist when n ≤ 5. But, when n ≥ 6, a 3-GDD (n, 3, 5, μ₁, 0) exists for some μ₁.

Theorem 17. For n ≥ 6, a 3-GDD (n, 3, 5, 3(n − 1)(n − 2)(n − 3)(n − 4)/4, 0) exists.

Proof. When n ≥ 6, taking $$ = (n − 3)(n − 4)/2 copies of the blocks of Theorem 15(a) together with 3(n − 1)(n − 2)/2 − n(n − 1) = (n − 1)(n − 6)/2 copies of the blocks of Theorem 15(b) give the blocks of 3-GDD (n, 3, 5, 3(n − 1)(n − 2)(n − 3)(n − 4)/4, 0).

Example 4. By Theorem 17, the following 3-GDDs exist:

•  

  3-GDD (6, 3, 5, 90, 0), 3-GDD (7, 3, 5, 270, 0), 3-GDD (8, 3, 5, 630, 0)

•  

  3-GDD (9, 3, 5, 1260, 0), 3-GDD (10, 3, 5, 2268, 0), 3-GDD (11, 3, 5, 3780, 0)

•  

  3-GDD (12, 3, 5, 5940, 0), etc.

Theorem 18. There exists a 3-GDD (n, 3, 5, (n − 1)(n − 2)(n − 3)(n − 4)/4, 0) for n ≥ 6 and n ≡ 2(mod 3).

Proof. If n ≡ 2(mod 3) and n ≥ 6, then both (n − 1)(n − 6)/2 and (n − 3)(n − 4)/2 are divisible by 3. Joining (n − 3)(n − 4)/6 copies of the blocks of Theorem 15(a) together with (n − 1)(n − 6)/6 copies of the blocks of Theorem 15(b) give the blocks of 3-GDD (n, 3, 5, (n − 1)(n − 2)(n − 3)(n − 4)/4, 0).

Example 5. By Theorem 18 above, the following 3-GDDs exist:

•  

  3-GDD (6, 3, 5, 30, 0), 3-GDD (7, 3, 5, 90, 0), 3-GDD (9, 3, 5, 420, 0)

•  

  3-GDD (10, 3, 5, 756, 0), 3-GDD (12, 3, 5, 1980, 0), 3-GDD (13, 3, 5, 2970, 0)

•  

  3-GDD (15, 3, 5, 6006, 0), etc.

Example 6. A 3-PBIBD (3, 3, 5, 0, 3, 4) with G₁ = {1,2,3}, G₂ = {a, b, c}, G₃ = {x, y, z} exists and the blocks (written vertically) of this design are as follows.

Example 7. A 3-PBIBD (3, 3, 5, 6, 3, 0) exists and the blocks of this design are obtained by union of every G_i with every two element set of G_j for i ≠ j, where i, j = 1, 2, 3.

Example 8. 3-PBIBD (3, 3, 5, 9, 3, 3) exists. Its blocks are obtained by union of every group with each singleton set of both remaining groups.

Remark 19. When n = 3, from the original necessary conditions in Theorem 7; r, b, λ₁ and λ₂, respectively, are integers if ∗₁: μ₁ +3 μ₂ ≡ (mod  6), ∗₂: μ₁ + μ₂ ≡ (mod  10), μ₁ ≡ 0 (mod  3), and μ₁ ≡ 0 (mod  3).

Thus, the values of μ₁ and μ₁ satisfying the necessary conditions are as follows: μ₁ ≡ 0(mod  3)∩∗₁∩∗₂ and μ₂ ≡ ∗₁∩∗₂.

For n = 3, taking all the possible combinations, the original necessary conditions are satisfied when

• (1)

  μ₁ ≡ 3(mod  30) and μ₂ ≡ 7(mod  10)

• (2)

  μ₁ ≡ 9(mod  30) and μ₂ ≡ 1(mod  10)

• (3)

  μ₁ ≡ 15(mod  30) and μ₂ ≡ 5(mod  10)

• (4)

  μ₁ ≡ 21(mod  30) and μ₂ ≡ 9(mod  10)

• (5)

  μ₁ ≡ 27(mod  30) and μ₂ ≡ 3(mod  10)

• (6)

  μ₁ ≡ 0(mod  30) and μ₂ ≡ 0(mod  10)

• (7)

  μ₁ ≡ 6(mod  30) and μ₂ ≡ 4(mod  10)

• (8)

  μ₁ ≡ 12(mod  30) and μ₂ ≡ 8(mod  10)

• (9)

  μ₁ ≡ 18(mod  30) and μ₂ ≡ 2(mod  10)

• (10)

  μ₁ ≡ 24(mod  30) and μ₂ ≡ 6(mod  10)

Example 9. 3-GDD (3, 3, 5, 9, 11) exists and its blocks are obtained by combining the blocks of Example 3 with two copies of the blocks of Example 6.

Example 10. 3-GDD (3, 3, 5, 6, 4) exists and its blocks are obtained by joining blocks of Examples 6 and 7.

Hence, we have 3-GDD (3, 3, 5, 6t, 4t), for t ≥ 1.

Theorem 20. There exists a 3-GDD (3, 3, 5, 9q + 3p, 11q − 3p) for p ≤ q.

Proof. Blocks of the design are obtained by taking p copies of 3-GDD (3, 3, 5, 12, 8) together with q − p copies of Example 9.

Corollary 21. There exists a 3-GDD (3, 3, 5, 9q + 3p + 6, 11q − 3p + 4) for p ≤ q.

Proof. Taking blocks of 3-GDD (3, 3, 5, 9q + 3p, 11q − 3p) together with blocks of Example 10 give 3-GDD (3, 3, 5, 9q + 3p + 6, 11q − 3p + 4).

Corollary 22. For all p ≥ 1, 3-(9, 5, 30p) exists.

Proof. Blocks of such design are obtained by taking p-copies of 3-GDD (3, 3, 5, 12, 8) together with 2p-copies of Example 9.

Corollary 23. There exists a 3-(9, 5, 30p + 15) for all p ≥ 0.

Proof. For p ≥ 0, blocks of 3-(9, 5, 30p + 15) are obtained by joining p-copies of blocks of 3-GDD (3, 3, 5, 12, 8), 2p + 1-copies of blocks of Example 9 and blocks of Example 10.

The next Theorem discusses the situation in which the existence of a 3^′-GDD (Definition 2) guarantees the existence of 3-PBIBD (Definition 4).

Theorem 24 (Block Complementation). Let k ≤ mn − 3. If a 3^′-GDD (n, m, k, Λ₁, Λ₂) exists, then a 3-PBIBD (n, m, nm − k, θ₁, θ₂, θ₃) also exists, where θ₁ = b − 3r + 3λ₁ − Λ₁, θ₂ = b − 3r + λ₁ + 2λ₂ − Λ₂, θ₃ = b − 3r + 3λ₂ − Λ₂. r, and λ₁&λ₂, respectively, are the replication number, number of first associate pairs, and number of second associate pairs in the 3^′-GDD.

Proof. Suppose a 3^′-GDD (n, m, k, Λ₁, Λ₂) exists.

Let ($$, $$) is a 3-GDD (n, m, k, Λ₁, Λ₂), where $$ is a mn set of points partitioned in to m groups of size n each, and $$ is a collection of k-subsets (blocks) of $$.

To Show ($$,{$$: $$}) is a 3-PBIBD (n, m, mn − k, θ₁, θ₂, θ₃).

This design has mn points partitioned in to m-parts (groups) of size n, and every block contains mn − k ≥ 3 symbols.

To show every triple of type,

• (i)

  (3,0) occurs in θ₁ = b − 3r + 3λ₁ − Λ₁ blocks

• (ii)

  (2,1) occurs in θ₂ = b − 3r + λ₁ + 2λ₁ − Λ₂ blocks

• (iii)

  (1,1,1) occurs in θ₃ = b − 3r + 3λ₂ − Λ₂ blocks

  • (1)

    To show every triple of type (3,0) occurs in θ₁ = b − 3r + 3λ₁ − Λ₁ blocks,

  •  

    Let (x, y, z) is triple of type (3,0) in ($$, $$). There are 3r − 3λ₁ + Λ₁ blocks containing x or y or z (containing at least one of them), which means b − 3r + 3λ₁ − Λ₁ blocks in ($$, $$) do not contain any one of the three elements.

  •  

    Therefore, in ($$, {$$: $$}), and (x, y, z) occurs in θ₁ = b − 3r + 3λ₁ − Λ₁ blocks.

  • (2)

    To show every triple of type (2,1) occurs in θ₂ = b − 3r + λ₁ + 2λ₂ − Λ₂ blocks. Let (x, y, z) is triple of type (2,1) in ($$, $$). There are 3r − λ₁ − 2λ₂ + Λ₂ blocks containing x or y or z (each block containing (x, y, z) contains one first and two second associate pairs).

  •  

    Thus, in ($$, {$$: $$}), there are θ₂ = b − 3r + λ₁ + 2λ₁ − Λ₂ blocks containing x, y, and z.

  • (3)

    Similarly, if (x, y, z) is triple of type (1,1,1) in ($$, $$), there are 3r − 3λ₂ + Λ₂ blocks containing x or y or z (and each block containing (x, y, z) contains three second associate pairs), and as a result there are θ₃ = b − 3r + 3λ₂ − Λ₂ blocks not containing the triple.

Thus, in ($$, {$$: $$}), there are θ₃ = b − 3r + 3λ₂ − Λ₂ blocks containing x, y and z.

Corollary 25.

• (1)

  If a 3^′-GDD (3, 3, 4, Λ₁, Λ₂) exists then 3-GDD (3, 3, 5, (9Λ₂ + Λ₁)/4, (11Λ₂ − Λ₁)/4) exists

• (2)

  If a 3-GDD (3, 3, 5, μ₁, μ₂) exists, then 3^′-GDD (3, 3, 4, (11μ₁ − 9μ₂)/5, (μ₁ + μ₂)/5) exists

Proof.

• (a)

  Suppose 3^′-GDD (3, 3, 4, Λ₁, Λ₂) exists

•  

  By Theorem 24, when m = 3 and k = 4, a 3-PBIBD (3, 3, 5, θ₁, θ₂, θ₃) exists, and n = 3 yields θ₁ = θ₂ = (9Λ₂ + Λ₁)/4 and θ₃ = (11Λ₂ − Λ₁)/4.

• (b)

  If a 3-GDD (3, 3, 5, μ₁, and μ₂) exists, then taking the complement of its blocks gives a 3-PBIBD (3, 3, 4, β₁, β₂, β₃), where β₁ = (11μ₁ − 9μ₂)/5, β₂ = β₃ = (μ₁ + μ₂)/5.

As a result of Corollary 25(b), we have the following examples of a 3^′-GDDs when k = 4.

Example 11. A 3^′-GDD (3, 3, 4, 6, 2) exists and its blocks are obtained by taking the complement of the blocks of Example 10.

Example 12. A 3-(9, 4, 6) exists. Its blocks are obtained by combining the complements of the blocks of Examples 9 and 10.

In general, taking the complements of the blocks of Corollary 23, the following result follows.

Corollary 26. For all p ≥ 0, a 3–(9, 4, 12p + 6) exists.