# 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?

• No, the answer is definitely the first one. I haven't seen any book where you use addition for the problem.
– user371838
Commented Dec 28, 2016 at 11:32
• The correct answer is $\binom{12}{4}\binom{8}{3}$. The answer $\binom{12}{4} + \binom{8}{3}$ would be correct if we were asked in how many ways can $4$ black marbles or $3$ green marbles be chosen? Commented Sep 14, 2018 at 10:05

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:

$$\prod_{i=1}^{k}n_{i}=n_{1}n_{2}....n_{k}$$