Putnam/Olympiad problem of the day

[http://math.harvard.edu/putnam(...) !!84u21X0KjW+ No.10617778]

[math]
\text{The 20 members of a local tennis club have scheduled exactly 14 two-person}
\\
\text{games among themselves, with each member playing in at least one game.}
\\
\text{Prove that within this schedule there must be a set of 6 games with 12 distinct}
\\
\text{players.}
[/math]

[http://math.harvard.edu/putnam(...) !!84u21X0KjW+ No.10617779]

Previous Thread >>10615013

[Anonymous No.10617784]

>math problem presented using sports analogy

[Anonymous No.10617786]

It's just a [math] (20, 14, 2) [/math] design

[Anonymous No.10617799]

>>10617778
This is pretty easy. Suppose there is a way of picking 5 games with 10 distinct players, then the remaining 10 players need to play in the 9 remaining games, so by the pigenhole principle, at least one of those 9 games has a pair of players who didn’t play in the 5 games, so there is a way of picking 6 games with 12 distinct players whenever there is a way of picking 5 games with 10 distinct players. Repeat for 4 games, 3 games, 2 games, and 1 game is an easy base case.

[Anonymous No.10617801]

>>10617784
it's to hide the fact it's an easy as shit olympiad combinatorics problem

[Anonymous No.10618202]

thinly veiled homework thread