$$ \begin{array} { l } { \text { In how many ways can one pick } n \text { marbles from } 3 n \text { marbles consisting of } n \text { indistinguishable } } \\ { \text { white marbles, } n \text { indistinguishable black marbles and } n \text { distinguishable coloured marbles. } } \\ { \text { Your solution should contain a precise explanation of your reasoning. } } \end{array} $$
I have tried to attempt this from working from the number of ways if they were all distinguishable and then minus-ing any repeats, but I am quite stuck.
To count the number of ways to pick $n$ marbles from the $3n$, we take cases on how many colored marbles are to be chosen. If your set of $n$ marbles is to include exactly $k$ colored marbles, there are $\binom{n}{k}$ ways to pick which colored marbles are included. Here, $k$ will range from $0$ to $n$. No matter the value of $k$, there are always at least $n-k$ white marbles available and at least $n-k$ black marbles available, because there are $n$ marbles of each type. Thus, by stars and bars, there are $n-k+1$ ways to choose how many white marbles are to be included in your chosen group of marbles. Overall, there are $(n-k+1)\binom{n}{k}$ choices of $n$ marbles such that $k$ are colored.
Summing over $k$ we see that the total number of choices is given by $$ \begin{split} \sum_{k=0}^n (n-k+1)\binom{n}{k} &= (n+1) \sum_{k=0}^n \binom{n}{k} - \sum_{k=0}^n \left( \frac{k \cdot n!}{k!(n-k)!} \right) \\ &= (n+1)2^n - n \sum_{k=0}^n \left( \frac{(n-1)!}{(k-1)!((n-1) - (k-1))!} \right) \\ & = (n+1)2^n - n \sum_{\ell = 0}^{n-1} \binom{n-1}{\ell} = (n+1)2^n - n2^{n-1} \\ &= 2^n + n2^{n-1}. \end{split} $$