In a set of cardinality 15, find 15 subsets, each of size 7, such that the intersection of any two would give a set with only 3 elements. The title in math form:

$|S|=15$ (The big superset)
$R=\{Q_0,Q_1,Q_2\dots Q_{14}\}$ (a set containing the 15 subsets)
$\forall n \in R, |n|=7$ (showing that each subset has 7 elements
$\forall (x_0, x_1) \in R\,\times R \:\:(x_0\neq x_1),\:\: |x_0\cap x_1| = 3$ (for all of the permutations that are possible ($x_0$ and $x_1$ cannot be the same because you can't have a pair with the same number twice)

Any help on this hard question would be very appreciated :)
Cheers!
 A: Hint: Add an element that is in all of the subsets. The problem then becomes

*

*Find a base set of $16 = 2^4 $ elements.

*Find 15 subsets, each of which have $8 = 2^3 $ elements.

*We want the intersection of any 2 of these subsets to give us $ 4 = 2^2 $ elements.


(Slight wishful thinking here)
These values are much nicer to work with, and the multiplicative nature suggests that we look at a certain vector space.
If the subsets were vector subspaces, then the intersection of subsets is the intersection of vector subspaces, which is another subspace. The index suggests that we've gone down 1 dimension.
So, we want to

*

*Start with a field of X elements,

*Create a vector space of dimension A

*Find 15 subsets of dimension B, whose intersection always gives us subsets of dimension C. (This might not be true in general)

*These have the common element 0, which is in every subspace.


And if you're still stuck:

 $ X = 2, A = 4$ is a natural choice.
 Consider the vector space  $ V = \{ 0, 1 \} ^ 4 $.


 This requires $B = 3$ and $ C = 2$.
 Thankfully, in a vector space of dimension 4, any 2 subspaces of dimension 3 will have an intersection of dimension 2.


 How do we create a subspace of dimension 3? We find the vector that it is perpendicular to.
 There are 15 non-zero $v \in V$.
 For every $ v \neq 0$, define $ T_v$ to be the set of vectors that are perpendicular to $v$.
 Show that $|T_v| = 8 $, since this is the subspace that is perpendicular to 1 non-zero vector.


 For every distinct $ u, v \neq 0 $, show that $ | T_u \cap T_v| = 4$, since this is the subspace that is perpendicular to 2 non-zero vectors.


 Now, remove the $0$ vector from everything.

