Number of ways of selecting all k-indexed identical items before all k+1 indexed identical items for all k from 1 to n Suppose we have n indices and we have a specific number of items allotted to this index. Say for 2 balls of colors Blue(B)[1], 4 of color Green(G)[2] and 2 of color Red[3] (I could've just assigned indices to the balls instead, but just to make this clearer) I want all the number of ways to select 1 indexed balls(Blue) before 2 indexed balls (G) before 3 indexed balls(R). Ofcourse, all balls of the same color are identical. These conditions have to be simultaneously satisfied in every picking. What is the approach?
Eg: If i have 2 balls of color B, 2 balls of color G, 1 ball of R:
Pickings are:
BGBGR,
BBGGR,
BGBGR 
 A: Some notation: we have $n$ colours and for each colour $i=1,\ldots,n,\;$ there are $b_i$ indistinguishable balls.
I take your meaning as follows: that the last ball of colour $i+1$ must follow the last ball of colour $i,\;$ for colours $i=1,\ldots,n-1,\;$ (i.e. all colours).
Without this restriction, the number of arrangements of the balls is the multinomial coefficient
$$M = \dfrac{\left(\sum\limits_{i=1}^{n} b_i\right)!}{\prod_{i=1}^{n}{b_i}!}.$$
Consider just the first two colours, $1$ and $2$. The last colour-$2$ ball will follow the last colour-$1$ ball in exactly $\dfrac{b_2}{b_1+b_2}$ of the above arrangements. To see this, it might help to ignore/remove the balls of all other colours and note that the probability of having a colour-$2$ ball last in the resulting sub-arrangement of balls of colours $1$ and $2$ is $\dfrac{b_2}{b_1+b_2}$.
So the number of arrangements where only colours $1$ and $2$ satisfy the criterion is $M\dfrac{b_2}{b_1+b_2}$.
Next we consider the balls of colour $3$ in these $M\dfrac{b_2}{b_1+b_2}$ arrangements. We see that, by similar reasoning, the last colour-$3$ ball will follow the last balls of both preceding colours in $\dfrac{b_3}{b_1+b_2+b_3}$ of them.
In general, in satisfying the requirement that the last ball of colour $i$ must follow the last ball of the previous colour adds a factor of $\dfrac{b_i}{b_1+\cdots+b_i}$.
Thus, the number of arrangements is
$$M = \dfrac{\left(\sum\limits_{i=1}^{n} b_i\right)!}{\prod_{i=1}^{n}{b_i}!} \prod_{i=1}^{n-1}{\dfrac{b_i}{b_1+\cdots+b_i}}.$$
$$\\$$
To illustrate this with your example, we have $n=3$ and $b_1=2,\; b_2=2,\; b_3=1$. So the number of arrangements is
$$\dfrac{\left(2+2+1\right)!}{2!2!1!} \dfrac{2}{2+2} \dfrac{1}{2+2+1}=3.$$
