Alena, Bára, Čeněk, and David jointly bought a tandem - a bicycle for two. They always set out on a ride in pairs. Everyone has already ridden with everyone else at least once, and no one else has ever ridden the tandem. Alena went on a ride eleven times, Bára twenty times, and Čeněk only four times.

Determine how many times David could have been on a ride at a minimum and at a maximum.
If Alena, Bára, and Čeněk each went with each other exactly once and the rest of the outings were spent with David, then David would be on the ride 29 times, because $(11-2)+(20-2)+(4-2)=29$. Twenty-nine is the highest possible number. Schematically, this solution is illustrated in the following image. (For example, the number next to the letter $\mathrm{B}$ indicates how many times Bára was on the ride in total, and the number on the connection BC expresses how many times Bára rode with Čeněk... )

![](https://cdn.mathpix.com/cropped/2024_04_17_cd0342d3a611cc15345bg-2.jpg?height=474&width=523&top_left_y=957&top_left_x=801)

For solving the rest of the problem, we want to ensure that Alena, Bára, and Čeněk ride together as many times as possible. Alena could ride with Bára a maximum of 9 times $(11-2=9)$. Čeněk could ride with both Alena and Bára a maximum of 2 times $(4-2=2)$, but he could not achieve this maximum with both at the same time, as he had to ride at least once with David. If Čeněk rode twice with Alena, the maximum number of rides between Alena and Bára could not be achieved, so Čeněk realized two rides with Bára. From this, it follows how many times David was on the ride: once with Alena and Čeněk, 9 times with Bára $(20-9-2=9)$, totaling 11 times $(1+1+9=11)$. For better orientation in the text, see the image:

![](https://cdn.mathpix.com/cropped/2024_04_17_cd0342d3a611cc15345bg-2.jpg?height=468&width=525&top_left_y=1933&top_left_x=800)

David could be on the ride a minimum of 11 times and a maximum of 29 times.