When to add and when to multiply with combination problems? Finding the number of ways a certain task can be done has always given rise to a few troubles for me. Issues seem to arise in questions where AND and OR rear they're heads. 
For example: 

Suppose a box contains 12 black marbles and 8 green marbles. How many ways can 4 black marbles and 3 green marbles be chosen?

To me, I feel as though the correct answer would be $\binom{12}{4}\times \binom{8}{3}$ i.e. for every 4 black marbles selected there are 3 possible green marbles which could be selected. 
However, the solution for this particular problem says that there are $\binom{12}{4} +  \binom{8}{3}$ ways to draw the marbles. 
Some searching online, however, suggests my solution is correct, while other sources provide the latter approach as a solution. 
What is the correct way to approach this question and ones similar to it?
 A: Sorry This is not a complete answer but i hope it can help you .
Rule of sum: 
" if we have (a) ways of doing something and (b) ways of doing another thing and we can not do both at the same time, then there are a + b ways to choose one of the actions."
if we can divide Event E into k events $E_{1},E_{2},....,E_{k}$ such that
$$n_{1}\space case \, to\, happen \,\,event \,\,E_{1}$$
$$n_{2}\space case \, to\, happen \,\,event \,\,E_{2}$$
$$.$$
$$.$$
$$.$$
$$n_{k}\space case \, to\, happen \,\,event \,\,E_{k}$$
and if don't happen two event together then the number of states to happen one of $E_{1},E_{2},....,E_{k}$ is equal to:
$$\sum_{i=1}^{k}n_{i}=n_{1}+n_{2}+....+n_{k} $$   

let $k\in N$ and $A_{1},A_{2},....,A_{k},k $  finite collection of
  pairwise disjoint sets;that's mean for each of i,j=1,2,....,k and $i\neq
 j$ if we have $A_{i}\cap A_{j}=\varnothing$ then we have :
$|\cup_{i=1}^{k}A_{i}|=|A_{1}\cup A_{2}  \cup.....\cup
 A_{k}|=\sum_{i=1}^{k}|A_{i}| $

Rule of product:
"if there are (a) ways of doing something and (b) ways of doing another thing, then there are a · b ways of performing both actions."
if we can divide Event E into k Consecutive events $E_{1},E_{2},....,E_{k}$ such that:
$$n_{1}\space case \, to\, happen \,\,event \,\,E_{1}$$
$$n_{2}\space case \, to\, happen \,\,event \,\,E_{2}$$
$$.$$
$$.$$
$$.$$
$$n_{k}\space case \, to\, happen \,\,event \,\,E_{k}$$
then the number of states to happen E (first $E_{1}$,then $E_{2},....,$ at the end $E_{k})$ is equal to:
$$\prod_{i=1}^{k}n_{i}=n_{1}n_{2}....n_{k}$$
