# Finding the number of ways of arranging $2n$ white and black balls each such that no $n$ consecutive white balls are together

The question is:

Find number of ways of arranging $$2n$$ white and $$2n$$ black balls such that no $$n$$ consecutive white balls are together.

What I did was to arrange the black balls and number the $$2n+1$$ gaps between them as $$x_i$$ where $$1\le i\le 2n+1$$ and now use the relation: $$\sum_{i=1}^{2n+1}x_i=2n$$ where $$0\le x_i\le n-1$$ and $$x_i$$ denotes number of white balls in the $$i^{th}$$ gap.

This yields the solution for the number of ways as the coeff. of $$x^{2n}$$ in $$(1+x+x^2+...+x^{n-1})^{2n+1}$$

This is where I'm having a problem. How do I calculate this coefficient?

Any help or alternate methods would be appreciated.

• I suppose you could write $(1+x+x^2+...+x^{n-1})^{2n+1}=(1-x^n)^{2n+1}(1-x)^{-2n-1}$ and then use series expansion/product, not sure if it yields anything nice.
– Sil
Sep 6, 2020 at 9:42
• @Sil I tried it but couldn't go too far. Sep 6, 2020 at 9:44
• The number with a group of $2n-k$ white balls is $2{2n+k-1\choose k}+(2n+k-1){2n+k-2\choose k}$ as the group needs a black ball at either end unless the group is at one end. Sep 6, 2020 at 10:24

To finish where you left off, to get the coefficient of $$x^{2n}$$ we write $$(1+x+x^2+...+x^{n-1})^{2n+1}=(1-x^n)^{2n+1}(1-x)^{-2n-1}.$$ Now we can use Binomial theorem and Binomial series expansion to get $$\left(\sum_{i=0}^{2n+1}(-1)^i\binom{2n+1}{i}x^{in}\right) \cdot \left(\sum_{j=0}^{\infty}(-1)^j\binom{-2n-1}{j}x^{j}\right).$$ Since we are interested in coefficient of $$x^{2n}$$, we must have $$in+j=2n$$, and so $$j$$ is divisible by $$n$$. Thus only possible choices for $$(i,j)$$ are $$(2,0)$$, $$(1,n)$$ and $$(0,2n)$$ and the coefficient is $$(-1)^2\binom{2n+1}{2}(-1)^0\binom{-2n-1}{0}+ (-1)^1\binom{2n+1}{1}(-1)^n\binom{-2n-1}{n}+\\ (-1)^0\binom{2n+1}{0}(-1)^{2n}\binom{-2n-1}{2n}$$ which simplifies to $$\bbox[#ffd,10px]{\binom{4n}{2n}-(2n+1)\left[\binom{3n}{n}-n\right]}.$$ Here we have used also $$\binom{-2n-1}{n}=\binom{3n}{n}(-1)^n$$ and $$\binom{-2n-1}{2n}=\binom{4n}{2n}$$, which follow directly from definition of generalized binomial coefficient $$\binom{\alpha}{k}=\frac{\alpha(\alpha-1)\cdots(\alpha-k+1)}{k!}$$.