Answer to Question #254865 in Discrete Mathematics for Unknown354520

Question

Fourteen chess players a, b, c, d, e, f, g, h, i, j, k, l, m, n are to be paired so they can play 7 games starting at the same time. The only pairs allowable are the following: {c,f}, {a,d}, {b,g}, {e,h}, {a,i}, {a,j}, {a,k}, {a,l}, {a,m}, {a,n}, {b,h}, {b,j}, {b,k}, {b,l}, {b,m}, {b,n}, {c,i}, {c, j}, {c,k}, {c,l}, {c,m}, {c,n}, {d,h}, {d,i}, {d,j}, {d,k}, {d,l}, {d, n}, {e,h}, {e,i}, {e,k}, {e,l}, {e, m}, {e, n}, {f,i}, {f,j}, {f, k}, {f,l}, {f,m}, {f,n}, {g,i}, {g,j}, {g,k}, {g,l}, {g,m}, {g,n}, {h,i}, {h,j}, {h, m}. Find a possible pairing.

Answer 1

number of pairs allowable: 49

Total number of possible pairs:

$n=C^2_{14}=\frac{14!}{2!12!}=91$

 possible pairing:

$49/91=7/13$