Probability of drawing n balls of same color in case of groups of identical balls . A bag contain n identical red balls ,2n identical black balls and 3n identical white balls. if probability of drawing n balls of same color is greater than  or equal to 1/6 , then the minimum number of red balls in the bag is equal to ?
Solution from text book : we have n identical red balls,2n identical black balls and 3n identical white balls.
Total number of selection of n balls from n,2n and 3n identical white balls , X1+X2+X3 = n.
= Coefficient of $x^{n}$ in$(1-{X^{n+1}}) * (1-{X^{2n+1}}) *(1-{X^{3n+1}}) *(1-{X})^{-3} $
= Coefficient of $X^{3}$ in $(1-{X})^{-3} $ = ${n+3-1\choose 3-1}$ = ${n+2\choose 2}$
     **Favorable case = 1 + 1 +1 = 3.**
     
     $$
                            Required probability =  \frac { 3 }{${n+2\choose 2}$}
     $$ 
                           which will be greater than 1/6.
     
     I have two doubts in this solution (a) how the solution of   X1+X2+X3 = n.
        = Coefficient of $x^{n}$ in$(1-{X^{n+1}}) * (1-{X^{2n+1}}) *(1-{X^{3n+1}}) *(1-{X})^{-3} $
     
    (b) Favorable case = 1 + 1 +1 = 3.

 Please help me out form these two doubts !

 A: I assume that the question is meant to read that we have $n$ red, $2n$ black and $3n$ white balls, that we draw $n$ of the balls (this crucial piece of information was left out of what you wrote), and we ask whether or not all of the balls we drew were all the same color.
Yes, the balls themselves may in reality be indistinguishable apart from color, however that is not conducive to calculations.  When we talk about randomly pulling balls without further clarification it is always meant to be assumed that each physical ball is equally likely to have been pulled, not that each color is equally likely to be pulled.
We make the modification of the problem statement then to read that the balls are in fact distinctly labeled.  We are perfectly within our rights to do so and it does not change the probabilities involved.  This allows us to work with an equiprobable sample space and allows us to then calculate probabilities using counting techniques.
The cited answer from the book makes the critical error of working with the original sample space which is not equiprobable and incorrectly using counting techniques on this not-equiprobable sample space leads to incorrect results.
Now... as for the correct probabilities, having treated these balls all as distinct but grouped according to color, we can apply the hypergeometric distribution.  We could either have been in a favorable outcome of all balls drawn being the same color because of them all being red, because of them all being black, or because of them all being white.  The probability of this happening is going to be:
$$\dfrac{\binom{n}{n}\binom{5n}{0}}{\binom{6n}{n}}+\dfrac{\binom{2n}{n}\binom{4n}{0}}{\binom{6n}{n}}+\dfrac{\binom{3n}{n}\binom{3n}{0}}{\binom{6n}{n}} = \dfrac{1+\binom{2n}{n}+\binom{3n}{n}}{\binom{6n}{n}}$$
We ask for what values of $n$ this will be greater than or less than $\frac{1}{6}$.  For that, plug it into a calculator like wolframalpha and consult the table.

I subtracted $\frac{1}{6}$ so I could more easily see which of the probabilities were larger than $\frac{1}{6}$ (and so result in a positive number after the subtraction) compared to which are less than $\frac{1}{6}$ (and so result in a negative number after the subtraction).
We see that for $n=1$ and $n=2$ the probability of drawing all the same color balls will be greater than $\frac{1}{6}$ and we can show that for $n\geq 3$ the probability of drawing all the same color will be less than $\frac{1}{6}$
