Combinatorics distribution problem indistinguishable items in distinguishable boxes In how many ways can you put $10$ identical gold coins into four colored boxes so that at least $1$ goes into the blue box, at least $1$ into yellow, at most $2$ into red and at least $3$ into green?
The way I solved this was by writing down all the restrictions, 
$$B \geq 1, \space G \geq 3, \space Y \geq 1, \space R \leq 2$$ Where each letter corresponds to the first letter of each colour box.
Dividing the problem into 3 cases where there's no coin in red box, 1 coin and 2 coins in red box seems to be the most logical, giving the following results using bars and stars:
C1: R box has zero coins
$\binom{3 + 5-1}{5} = \binom{7}{5}$
C2: R box has one coins
$\binom{3 + 4 - 1}{4} = \binom{6}{4}$
C3: R box has two coins
$\binom{3 + 3 - 1}{3} = \binom{5}{3}$
Giving us a final result of $$\binom{7}{5}+\binom{6}{4}+\binom{5}{3}$$
I wanted to know if this is the correct way of solving a problem like this? What if we had bigger numbers including a box having at most say a 1000 coins, would we need to make a 10000 cases? what if there's multiple at most restrictions? like 2 boxes have to have at most 3 gold coins? 

There is an easier way to solve this problem which generalizes better to when there are several boxes which may have at most $N$ coins, where $N$ is large. I will illustrate with your example. 
First, count the number of ways to put all the coins into the boxes without the $R\le 2$ restriction. This is $\binom{5+4-1}5$, beacuse you are putting $5$ coins into $4$ boxes. Next, subtract out the "bad" distributions where the $R$ box has more than $2$ coins. This is equivalent to counting distributions satisfying $\{B\ge 1,G\ge 3, Y\ge 1, R\ge 3\}$, the number of which is $\binom{2+4-1}2$. The final answer is $\binom{8}5-\binom{5}2$.
What if you had two boxes with an upper limit? Say you have $15$ balls, and you still have all four boxes with their restrictions from before, but now there is also a purple box which can have at most $5$ balls. 


*

*Start by ignoring both the upper limit restriction, counting $\{B\ge 1,G\ge 3,Y\ge 1\}$. The number is $\binom{10+5-1}{10}$. 

*Subtract out the distributions where the $R$ box has more than $2$ balls. This is counting $\{B\ge 1,G\ge 3,Y\ge 1, R\ge 3\}$ The count is $\binom{7+5-1}{7}$

*Subtract out the distributions where the $P$ box has more than $5$ balls. This is counting $\{B\ge 1,G\ge 3,Y\ge 1, P\ge 6\}$ The count is $\binom{4+5-1}{4}$

*Now, the cases where both the $R$ and $P$ boxes have been subtracted out twice, so these cases must be added back in. We are adding back in placements satisfying $\{B\ge 1,G\ge 3,Y\ge 1, R\ge 3,P\ge 6\}$, the number of which is $\binom{1+5-1}{1}$.
The final answer is
$$
\binom{14}{10}-\binom{11}{7}-\binom{8}{4}+\binom{5}1
$$
In general, if there are many boxes with an upper limit, you have to use the Principle of Inclusion-Exclusion to subtract out the bad distributions.
