General counting problem. Not sure how to proceed. Mr. Popular has seven friends. He wants to count the number ways that he can invite a different subset of 3 of his friends for a dinner on 7 straight evenings s.t. each pair of friends are together for just 1 dinner.
I tried to break this problem into smaller ones. Here's what I have so far: I created a table to show one possible collection of his triples s.t. no 2 friends are paired more than once. 
$$
 \begin{pmatrix}
  1 & 1 & 1 & 0 & 0 & 0 & 0   \\
  1 & 0 & 0 & 1 & 1 & 0 & 0   \\
  1 & 0 & 0 & 0 & 0 & 1 & 1   \\
  0 & 1 & 0 & 1 & 0 & 0 & 1  \\
  0 & 1 & 0 & 0 & 1 & 1 & 0  \\
  0 & 0 & 1 & 1 & 0 & 1 & 0  \\
  0 & 0 & 1 & 0 & 1 & 0 & 1  \\
 \end{pmatrix}
$$
It doesn't look like I can use the LaTeX command "\bordermatrix", so please bare with me as I explain what the matrix means. The columns should be labeled 1 through 7 (representing each evening). The rows I have labeled as $a, b, c, d, e, f, g$ (friends). A "$1$" just means they're selected. So the first night (first column), friends $a, b,$ and $c$ are having dinner with Mr. Popular.
So if I could figure out how many collections there are, then I could multiply that by $7!$, where 7! is the arrangements from 1 collection. I don't think I should try to figure out how many different collections there are by hand, but I'm stuck otherwise. So what I think the answer should be is
$$
(\text{number of collections}) \cdot 7!
$$
 A: To supplement to the answer by @DouglasSStones, "the" projective plane of order 7 (seven points and seven "lines" of three points each) is known as the Fano plane:

We can sketch an argument as to why this "block design" is unique up to isomorphism (maps preserving lines and incidence).  Given seven points we ask for blocks ("lines") each containing three points, that any pair of points determines a line (there exists a unique block containing that pair), and that any two lines intersect in exactly one point.
It follows that two distinct blocks are either disjoint or share exactly one point.  If we sum the count of points over all lines, we get three times the number of lines.  But each point belongs to three lines, as removing that point from the lines incident to it partitions the remaining six points into three disjoint pairs.  So three times the number of lines also counts each point three times, proving the number of lines equals the number of points: seven.
Since a given point $X$ belongs to just three lines, there are exactly four blocks that miss point $X$.  For such a point $X$, fix three lines that each meet $X$ and two more points, say $\{X,A_1,A_2\}$, $\{X,B_1,B_2\}$, $\{X,C_1,C_2\}$.  We need to prove the six points other than $X$ can be organized into four blocks in essentially just one way, i.e. up to a relabeling of points.
Each of those four blocks contains one point (not $X$) from each of those three lines.  Swap subscripts of labels as necessary so that $\{A_1,B_1,C_1\}$ is one of them.  As this latter block must meet each of the other final three blocks in exactly one point, they are forced:
 $\{A_1,B_2,C_2\}$, $\{A_2,B_1,C_2\}$, $\{A_2,B_2,C_1\}$.
This sketch also accounts for the automorphism group having order 168.  We can map point $X$ to a point $Y$ in the Fano plane in seven ways.  The three lines meeting $X$ can be mapped to the respective three lines meeting $Y$ in $6 = 3!$ ways.  Finally if we pick one of the four lines not meeting $Y$ to be the image of $\{A_1,B_1,C_1\}$, the rest of the correspondence is forced.  Hence one gets $7\cdot 6\cdot 4 = 168$ possible automorphisms.
