v = 15, b = 35, r = 7, k = 3, λ = 1 (see details here and here)

Each configurations consists of a plane of the geometry PG(3,2); if we denote this configuration by 7₃, the first formula is then {7₃}¹⁵. There is only one 7₃ configuration. Each pair of planes meets in a line: and there are 455 selections of 15 planes, 3 at a time. Of these, 35 are accounted for by the fact a given line has 3 planes through it. As for the remaining 420, let 3 planes be A,B,C. Define (AB) to be the line of intersection of A and B, and define (AC) and (BC) likewise. A line not contained in a given plane must meet that plane in a point, so (AB) meets C in a point, and this point must lie on (AC) and (BC) as well (since it lies on all 3 planes, it must lie on the intersection of any pair of them). So the remaining triples of configurations each consist of 3 lines through a given point. (This means that there should be 420/15 = 28 such triples through each point. Through a given point, there are 7 ways of choosing the first line, 6 ways of choosing the second, and 4 ways of choosing a third line not in the plane defined by the first two. Since these 3 lines could have been chosen in 6 ways, the number of distinct selections is 7.6.4/6 = 28, as required.) So the intersection formula is 3³⁵7⁴²⁰. All the formulae are clearly the same both ways.