Couldn't give a title 2

Each cricket match will require at least one ball, so we just to need to determine the number of ways of distributing the $3$ "extra" (identical) balls amongst the $15$ matches. With $a_{k}$ representing the number of extra balls required for match $k,$ the equation we must solve is

$\displaystyle\sum_{k=1}^{15} a_{k} = 3$ for non-negative integers $a_{k}, 1 \le k \le 15.$

This is a stars and bars calculation, which according to Theorem two in the link has solution

$\dbinom{3 + 15 - 1}{3} = \dbinom{17}{3} = \boxed{680}.$