A few years ago, the National Football League had two conferences, each with 13 teams. It was decided by the league office that each team would play a total of 14 games, 11 of which were to be with teams in their own conference and the other 3 games outside their conference. Show that this is not possible.

Question

hero · Answer

Is it too difficult?

anonymous · Answer

I'm not that good with sports ..

hero · Answer

What the....? You don't have to think of it as sports....

hero · Answer

Everyone must be stumped

anonymous · Answer

jk about the sports :) 
there are a few distractions going on, making me slower than usual.

hero · Answer

Oh okay

anonymous · Answer

Just trying to get an initial direction...
Is this question from the discrete math class ?
And somehow there is a graph representation here for the problem , in which we need to prove there is no euler path etc.  ?

hero · Answer

I already figured out the answer. I can't believe you got stumped

anonymous · Answer

Believe it, try to at least.

hero · Answer

The solution is to list the first thirteen teams as A - M. If you do that you will see that the first 12 teams can indeed play each other 11 times. But Team M will be left with no one to play in their conference because all of the spots will be filled

studybird123 · Answer

:)