Problem:
The $2013-14$ season was a short-lived ray of hope in an otherwise long dark night for the world’s greatest football team. The team played $38$ league games and the main contributing factor to the success was Luis Suarez $31$ goals.
$(i)$ Suarez was suspended for the first five games of the season after having bitten a Chelsea player the previous spring.
$(ii)$ The most goals he scored in a game was four, and this he achieved once.
$(iii)$ He scored two goals on six occasions.
How many possibilities does this leave for his goalscoring record ?
Solution:
Suarez played $33$ games.
There are $\binom{33}{1}$ choices for the game in which he scored four goals.
Then there are $\binom{32}{6}$ choices for the six games in which he scored two goals each.
This leaves $26$ games and we know that
$(a)$ He scored $0,1$ or $3$ goals in each of these games.
$(b)$ He scored a total of $31−4−2 \cdot 6 = 15$ goals in these $26$ games.
We seek the number of solutions to
$x_1 + \cdots + x_{26} = 15, \:\:\:\: x_i \in \{0,1,3\}$
We do a case-by-case analysis:
Case $1$: He scored $3$ goals five times and $0$ goals every other time. There are $\binom{26}{5}$ choices for the goalscoring games.
Case $2$: He scored $3$ goals four times, $1$ goal three times and $0$ goals otherwise. There are $\binom{26}{4}\binom{22}{3}$ choices for the goalscoring games.
and so on...
Putting it all together, using the multiplication and addition principles, the number of possibilities for his goalscoring record is
$$\binom{33}{1}\binom{32}{6}\left[\binom{26}{5} + \binom{26}{4}\binom{22}{3} + \binom{26}{3}\binom{23}{6} + \binom{26}{2}\binom{24}{9} + \binom{26}{1}\binom{25}{12} + \binom{26}{15} \right]$$
$$\approx 2.5521 \times 10^{16}$$
Is there are more efficient way to solve this problem and other similar problems where the constraints can be even more challenging? E.g. how would you use generating functions which is suggested in the comments?
You can do this using inclusion-exclusion. The number of ways of distributing $15$ goals over $26$ games such that $2$ goals were scored in $k$ particular games and at least $4$ goals were scored in $l$ particular games is
$$ \binom{15-2k-4l+26-k-1}{26-k-1}\;, $$
so the desired count is
$$ \binom{33}1\binom{32}6\sum_{k=0}^7\sum_{l=0}^3(-1)^{k+l}\binom{26}k\binom{26-k}l\binom{40-3k-4l}{25-k}=\binom{33}1\binom{32}6\cdot853423740\;, $$
in agreement with your calculation.