Let K be the complete oriented graph on the finite set of vertices A. A family G="{Ga:a∈A} of spanning subgraphs of K is an orthogonal cover provided every arrow of K occurs in exactly one Ga and for every two elements a,b∈A, the graphs Ga and Gopb have exactly one arrow in common.
Any two groups have exactly 10 members in common.
And the LBB last name and the son's name has exactly one letter in common in the same place (capitalized above).
It is proved in this paper that for any v≡9 (mod 18), v⩾45, there exists a pair of Kirkman triple systems of order v having exactly k triples in common if and only if k∈{0,1,…,tv−6,tv−4,tv}, with the possible exception k=tv−4.
The line graph (mathcal{L}(G)) of a graph G is the graph whose vertices are the edges of G, with two vertices in (mathcal {L}(G)) adjacent whenever the corresponding edges in G have exactly one vertex in common.
Denote by (X^{0}(n,t)) the n-vertex cactus consisting of t triangles and (n-2t-1) pedgest edgesuchch that triangles and pendent edges have exactly one vertex in common.
That theory is in fact the theory of inferences of a very specific sort: inferences with two premises, each of which is a categorical sentence, having exactly one term in common, and having as conclusion a categorical sentence the terms of which are just those two terms not shared by the premises.
When your train leaves the station, the first station you pass through should have exactly one initial letter in common with the station you just left.
It wasn't a performance, exactly (he had nothing in common with contemporaries, such as Gilbert and George).
Whether it's the "fat, coked-up, Tory MP dying on a rent boy" face or the "38-year-old Glaswegian Tramp" face, they have all had exactly the same thing in common – none of them are even slightly funny.
