Solving a combinatorial problem using partition I was given the next question: How many options are there to hand to two people 4 green apples, 3 red apples and 5 yellow apples.
I thought about using the partition method by multiplying each case. Is it the right way of looking at the problem and solving it? 
I'm sorry, English isn't my first language and I couldn't find the right definition of partition. What I meant by it, is the number of the options of selecting k items out of n without having any meaning to the order of the selected item  and allowing repetition
 A: Hint:
How many ways can you distribute only the red apples (ignoring that the other types of apples exist)?
How many ways can you distribute only the green apples (ignoring that the other types of apples exist)?

How many ways can you then distribute the red and green apples?
Additional Hint: The multiplication principle seems helpful here
Continue on then to solve the full problem in a similar fashion
A: At the risk of beating a dead horse, the problem is oversimplified by the fact that there are only $2$ people to spread the fruit around. So it would make more sense to look at it without the constraint of just two people.
But to avoid algebra initially, we can increase the number of people to $3$, and set the fruit basket to $4 \text{ green apples: } \color{green}{\huge \bullet}$; $3 \text{ red apples: } \color{red}{\huge \bullet}$; and $5 \text{ yellow apples: } \color{yellow}{\huge \bullet}$.
Distributing the $4$ (undistinguishable) green apples among three (distinguishable) people (Ann, Paul and Ted) would be equivalent to dividing the apples into three labeled boxes, represented by vertical bars. So for the example where the green apples get divided as...
$$\underset{Ann}{\underbrace{\color{green}{\huge \bullet}}}\quad\quad\underset{Paul}{\underbrace{\color{green}{\huge \bullet}\quad\quad\color{green}{\huge\bullet}}}\quad\quad \underset{Ted}{\underbrace{\color{green}{\huge \bullet}}}$$
the representation is:
 $$\huge \color{green}{ \bullet}\quad\vert\quad\color{green}{ \bullet}\quad\color{green}{ \bullet}\quad\vert \quad \color{green}{ \bullet}$$
The twist in the procedure is to consider that all the apples may go to one person, for example, to Ted:
$$\huge\vert\quad\vert\quad \color{green}{ \bullet}\quad\quad\color{green}{ \bullet}\quad\quad\color{green}{ \bullet}\quad \quad \color{green}{ \bullet}$$
or split $3$ for Ann, $1$ for Ted, and none for Paul:
$$\huge \color{green}{ \bullet}\quad\quad\color{green}{ \bullet}\quad\quad\color{green}{ \bullet}\quad\vert\quad\vert\quad \quad \color{green}{ \bullet}$$
Hence, for $k=3$ people, we just need $k-1=2$ separators (bars) to split the $4$ green apples. Therefore there are $n =4$ apples and $K-1 =2$ separators arranged as $n + k -1 =6$ "objects" or positions. As soon as the $n=4$ positions occupied by the apples are chosen, the rest of the positions will correspond to the two separators. Or we can choose first the positions occupied by the $k-1$ separators, likewise determining the separation of the apples. Hence the overall number of ways to separate the green apples is:
$$\binom{n+k-1}{n}=\binom{n+k-1}{k-1}=\binom{6}{4}$$
As pointed out before, the same would be done for the red and yellow apples.

For context this post uses the stars and bars method:

For any pair of positive integers $n$ and $k$, the number of
   $k$-tuples of non-negative integers whose sum is $n$ is equal to the
   number of multisets of cardinality $k − 1$ taken from a set of
   size $n + 1.$
Both numbers are given by the multiset number $\displaystyle
 \left(\!{\tbinom {n+1}{k-1}}\!\right) $, or equivalently by the
   binomial coefficient $\displaystyle {\tbinom {n+k-1}{k-1}}={\tbinom
 {n+k-1}{n}}$ or multiset number $\displaystyle {\big (}\!{\tbinom
 {k}{n}}\!{\big )}.$

