Counting the cup game

The balls could be place in a line that is parallel to one of the triangle's sides.

The number of ways this could happen is:

$\begin{aligned} 3\sum\limits_{k=3}^{6}\binom{k}{3} &= 3\binom{7}{4} \\ &= 105 \end{aligned}$

This sum is computed using the hockey stick identity .

The balls could also be placed in a line that is perpendicular to one of the triangle's sides.

The number of ways this could happen is $3\times 3=9.$

The total number of ways she could win is $105+9=\boxed{114}.$

Another way to compute this is to get the $7th$ value from a table A194131 : $1216$ . Then subtract it from the number of combinations of 3 points on the lattice $\binom{21}{3} - 1216 = \boxed{114}$ .

Notice that there are 2 cases for where the line connecting the balls can be. Case I: It goes through consecutive balls and Case II: It goes through some “empty space” not touched by any balls. For case I it is simply Hockey Stick but applied to all 3 sides, so 3*7C4=105. For Case II we can take it step by step. If the first ball lands on one of the vertices of the triangle, there is only one way to have the other 3 balls such that they are collinear. If the 1st ball lands on one of the rows consisting of only 2 balls, there are 2 cases to satisfy a collinear arrangement of balls. Therefore in total there are 3 arrangements that work, but remembering that we must account for all 3 sides, we add 9 to 105 to get 114.