Count minimum no of races

Question

Five swimmers training together either swam in a race or watched the others swim. At least how many races must have been scheduled if every swimmer had opportunity to watch all of the others?

Answer 1

Claim:$\;$The minimum is $4$ races.

Label the swimmers as $1,2,3,4,5$.

Every swimmer must race at least once (in order to be watched), and must watch at least once.

Thus, there must be at least $2$ races.

Suppose there are exactly $2$ races.

• Then every swimmer must swim in one race, and watch the other.$\\[4pt]$
• Hence the race watched by swimmer $1$ must have swimmers $2,3,4,5$, else swimmer $1$ won't get to watch them.
• But then swimmers $2,3,4,5$ must be watchers for the other race, contradiction, since they never get to watch each other.

Thus, $2$ races is not enough.

Next suppose there are exactly $3$ races.

• Since everyone must watch at least one race, some race must have at least $2$ watchers.$\\[4pt]$
• Without loss of generality, suppose swimmers $1,2$ are among the watchers for race $1$.$\\[4pt]$
• Since swimmer $1$ must watch swimmer $2$, and swimmer $2$ must watch swimmer $1$, each of swimmers $1,2$ races only once.$\\[4pt]$
• But then the race with swimmer $1$ must have everyone else watching, which means swimmer $1$ is the only swimmer for that race.$\\[4pt]$
• Similarly, the race with swimmer $2$ must have everyone else watching, which means swimmer $2$ is the only swimmer for that race.$\\[4pt]$
• It follows that race $1$ must have swimmers $3,4,5$, else swimmers $1,2$ won't get to watch everyone else.$\\[4pt]$
• But then swimmers $3,4,5$ never get to watch each other, contradiction.

Thus, $3$ races is not enough.

But $4$ races can work.

For example, using the $4$ races

• $2,3$
• $4,5$
• $1,3,5$
• $1,2,4$

everyone gets to watch everyone else.