Are there an infinite set of sets that only have one element in common with each other?

In a card game called Dobble, there are 55 cards, each containing 8 symbols. For each group of two cards, there is only one symbol in common. (The goal of the game being to spot it faster than the other players, which is not the point of my question).

If I translate that to mathematical language, I would say that:

  • $S = [S_1, S_2, …, S_{55}]$.
  • $S_n = [n_1, n_2, …, n_8]$.
  • For $S_n, S_m \in S$ there is one and only one $n_a = m_b$

My double (dobble) question is:

  • Are there a finite or infinite number of sets and elements that allows such a property? I know there is one more with 30 sets containing 6 elements each (because of Dobble Kids, a lighter version of the game).
  • How can I calculate the number of sets, the number of elements in the sets, how many different elements there are in all the sets and which elements go in which sets? Is there a formula or is it simply a step-by-step try and fail method?


I realise that having sets like {1, 2, 3, 4}, {1, 5, 6, 7}, {1, 8, 9, 10}, … answers the question (with 1 being the only element in common in each set). There is one more restriction:

  • Each element used in the sets must appear the same number of times (for example, in 7 given sets).

In the game, there are 50 symbols altogether. (55 cards, 8 symbols per card, 50 symbols altogether).

I have figured out a simple example with 4 sets of 3 elements (6 elements overall):
$$S_1 = [1, 2, 3], S_2 = [1, 4, 5], S_3 = [2, 5, 6], S_4 = [3, 4, 6]$$

Each element is present twice.

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This is not a complete explanation, but a summing up of my observations in the comments above.

Consider the projective plane ${\Bbb P}^2({\Bbb F}_q)$ over the field with $q$ elements ($q$ must be of the form $q=p^f$ for some prime $p$). Then the following facts follow more or less trivially from the definitions:

  • ${\Bbb P}^2({\Bbb F}_q)$ consists of $1+q+q^2$ points;

  • ${\Bbb P}^2({\Bbb F}_q)$ contains $1+q+q^2$ lines, each of them containing $q+1$ points;

  • every two lines meet at a single point;

  • every point is contained in exactly $1+q$ lines.

Thus, if we call “symbols” the points, and “cards” the lines we have a situation which is exactly thatdescribed in the question.

The problem is that the numeric data do not correspond: if we take $q=7$ so to match the $8$ symbols in each card, the number of cards and of symbols should be $1+7+7^2=57$.

Then, either you lost 2 cards [ 🙂 ], or I’m left clueless.

You can always define, for a set $I$ (which will be the index set, finite or not), something like this : Define $S_i = \{ \{i,j\} \, | \, j \in I \} $. (Since I chose the two-element subset $\{i,j\}$ instead of the ordered couple, order doesn’t matter here.) Clearly $S_i \cap S_j = \{\{i,j\}\}$ for $i \neq j$. But the sets $S_i$ are “just as big” as $I$ itself.

For instance, if $I = \{1,2,3,4\}$, then $S_1 = \{ \{1,1\} , \{1,2\}, \{1,3\}, \{1,4\} \}$, $S_2 = \{ \{2,1\} , \{ 2, 2\} , \{2,3\}, \{2,4\} \}$ and so $S_1 \cap S_2 = \{ \{1,2\} \}$.

Would that be a good example? (The way I came up with this is : imagine for some point in $\mathbb R^2$ on the $x=y$ line that some line goes vertical through this point and another one horizontally. Then for two points $(x_0,x_0)$ and $(x_1, x_1)$ on the $x=y$ line, the intersection of the vertical/horizontal lines of the first point and the vertical/horizontal lines of the second point are precisely $(x_0,x_1)$ and $(x_1,x_0)$. By choosing “non-ordered couples” I get unicity. You can produce an example with some large number of elements in common by simply choosing more “components” and considering ordered tuples instead.)

N = number of symbols per card;
S = total number of symbols;
C = total number of cards;

In stead of the maximum number of cards being S=N*N-N+1 and C=S, where (N-1) is prime, there is also a minimum number of cards, being S=(N+1)*N/2; C=N+1; and this is for all N>1.

SteeveDroz proved it already for N=3 with 6 Symbols and 4 Cards. where 2=(N-1) is prime too, that you do not need 7 Cards and 7 Symbols.

For N=4 there are 10 Symbols needed for the total of only 5 Cards;

For N=8 (the original Spot-It amount of Symbols per card) you need 9 cards and 36 Symbols, and not 57 Cards and 57 Symbols.
9 Cards with 8 Symbols each gives a total of 72 symbols on all cards. So each symbol occurs only 2 times in the pack of cards.

Example of deck of cards for N=4 and N=8