The question I am stuck on is 'how many ways can I choose $n$ balls from $3n$ balls, with $n$ indistinguishable black balls, $n$ indistinguishable white balls, and $n$ distinguishable coloured balls.'
I have been thinking about this problem for some time but can't get very far, I am used to questions about just distinguishable or indistinguishable objects, but I am not familiar with a problem involving both.
I think that there are $n+2$ 'colours' where within the colours all the balls are indistinguishable, but I don't really know where to go from there so any hints would be much appreciated.
Best Answer
I would do something like that: U have $n$ black and $n$ white balls which are indistinguishable and $n$ coloured balls which are distinguishable. U want to choose $n$ balls.
Let's assume that in our "way" we want to take k coloured balls. Then, there is exactly ${n \choose k}$ ways of choosing those coloured balls from $n$. Then we have $n-k$ balls missing and we need to choose those from black and white balls. But those are indistinguishable ! That means, we are only interested in number of (say) black balls in our "chosing". The number of black balls can vary from $j=0$ to $j=n-k$, which gives us $n-k+1$ ways of choosing the rest. Now, it is obvious that number of coloured balls ($k$) can vary from $0$, to $n$. So, the total number is given by: $$ T=\sum_{k=0}^n(n-k+1){n \choose k} $$
Simplier: $$ T=\sum_{k=0}^n(n+1){n \choose k} - \sum_{k=1}^nk{n \choose k}$$
$\sum_{k=0}^n(n+1){n\choose k} = (n+1)2^n$
$\sum_{k=1}^nk{n\choose k} = n2^{n-1}$ $\ \ \ \ \ $ (*)
So:
$$ T = (n+1)2^n - n2^{n-1} = n2^n + 2^n - n2^{n-1} = 2^{n-1}(2n + 2 - n) = (n+2)2^{n-1} $$
(*) Little explanation: RHS counts the number of ways of choosing a "captain" and his "team" from n people, where the number of people in "team" can be any from 0 to n-1. And we can see, that LHS counts the same, k - {number of people in a "team" + 1} ( cause we are to choose the captain from those k people, and the rest of them will form our "team" with k-1 people ( here k vary from 1 to n, then k-1 from 0 to n-1 and that is exactly what are we looking for)