How to calculate number of combinations for this problem? The problem is as follows:
You have an unlimited number of marbles and each marble is one of 16 different colours. You have to choose 6 marbles and order is irrelevant. How many different combinations of 6 marbles are there?
 A: Let's think that we have $16$ boxes of different colors and $6$ equal marbles to distribute among the boxes to color them.
Now let's call $m$ to the marbles and consider $16-1=15$ "separators" s. The problem comes down to finding the number of anagrams of the word with $15+6=21$ letters, $6$ of them $m$ and other $15$ are $s$.
The answer is $\dfrac{21!}{15!.6!}=\binom{21}{15}$ because you have to select where to place the separators within the 21 total spaces and then in the remaining places to place the marbles.
For example, the word $m-m-s-s-s-s-m-s-s-s-s-s-m-m-s - m -s -s -s -s-s $ returns $2$ marbles with color $1$, $1$ marble with color $5$, $2$ marbles with color $10$ and $1$ marble with color $11$.
A: Now I think one big problem with the approach in the comment and the solution that Vignolo proposed is that there isn't enough data to conclude there is infinity marble of each color.
For instance, if there is a condition that there is a maximum of 1 marble of type 1, then you would have to consider two cases, namely:

*

*There is no marble of type 1, then the answer to this case would be
\begin{align}
\binom{20}{14}
\end{align}

*There is exactly one marble of type 1, then the answer to this case would be
\begin{align}
\binom{20}{15}
\end{align}
To sum up, the answer to my hypothetical case should be
\begin{align}
\binom{20}{14} + \binom{20}{15}
\end{align}
Obviously, more intricate solutions would be required if there are two marbles of type I, or even worse, restraining conditions on 2 or 3 types of marbles.
