First, the number of balls in the box decrease by one at each step. Suppose $B(t)$ and $W(t)$ are the number of black and white balls present after $t$ steps. Since we start with $B(0)+W(0)=2731$ and $B(t+1)+W(t+1)=B(t)+W(t)-1$, we have that $B(2730)+W(2730)=1$. At this point we can no longer continue the process. So at this end, there is either a white ball or a black ball left.
Notice that the number of white balls present at any time is even. For instance, suppose we have just done step $t$. If we choose two black balls, then $B(t+1)=B(t)-1$ and $W(t+1)=W(t)$. If we choose a white ball and a black ball, then $B(t+1)=B(t)-1$ and $W(t+1)=W(t)$. If we choose two white balls, then $B(t+1)=B(t)+1$ and $W(t+1)=W(t)-2$. Necessarily, since $W(0)$ is even, it stays even throughout the process.
Hence $W(2730)=0$ and $B(2730)=1$. Even though what happens throughout the process is random, by parity, the process must end with only one black ball left.
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)
Best Answer
Let $(x,y)$ represent selecting a black ball labeled $x$ and a white ball labeled $y$.
Think of the possible valid ways to select two balls - either $|x-y|=0$ or $|x-y|=1$. This means that we can draw either $(x,x)$, $(x,x+1)$, or $(x,x-1)$.
Now for a crucial fact: it is impossible to draw two pairs of balls $(x,x+1)$ and $(x+1,x+2)$ and still meet the criteria given in the problem. If you were to do this, then there would be a black $x+2$ ball and a white $x$ ball still in the mix without any valid partners. You could argue that you could still pick $(x+2,x+3)$, for example, but this would then leave you in the same situation but now with a black $x+3$ and a white $x$.
From this, we get the following:
From here, let's call a drawing of the form $(x,x)$ to be a match, and a pair of drawings of the form $(x,x+1)(x+1,x)$ a switch, notated as $S_x$. We can show easily that the maximum number of switches possible is 3.
We can now say that a switch is even if $x$ is even and a switch is odd if $x$ is odd. It should be clear that we cannot have a drawing that includes both $S_x$ and $S_{x+1}$.
With all this in mind, it's easy to see which drawings are possible, and in fact, the list may be manually exhausted. Let's notate a drawing as a string of switches with the understanding that if a ball is not mentioned then it has a match. Our possible valid drawings are:
If the order that the pairs of balls are selected does not matter, then there are simply 19 different ways to pick the balls so that the criteria are met. If the order does matter, then there are $19\times 7!=95760$ ways.