So I'm completely new to this, and I have a very basic understanding of how this works. Here is my best attempt at trying to prove this.
${2n \choose n}$
=$\frac{(2n)!}{(n)!(2n-n)!}$
=$\frac{(2n)!}{2(n)!}$
and then
${2n-1 \choose n}$
=$\frac{(2n-1)!}{(n)! (2n-1-n)!}$
=$\frac{(2n-1)!}{(n)! (n-1)!}$
I have no idea if this is the method to prove this problem, or even if I have done it correctly up to this point. I hit a block in trying to get the ${2n \choose n}$ equivalent to the ${2n-1 \choose n}$.
I am really at a beginner level here, so if you could explain your steps it would be great!
Best Answer
First of all you made a mistake when you wrote that $(n)!(n)!=2(n)!$. It should be $(n!)^2$. So $\binom{2n}{n}=\frac{(2n)!}{(n!)^2}$.
Next, you got that $\binom{2n-1}{n}=\frac{(2n-1)!}{(n)!(n-1)!}$. Now just multiply and divide the expression by $2n$. You will get $\frac{(2n)!}{2(n!)^2}$. Hence $2\binom{2n-1}{n}=\frac{(2n)!}{(n!)^2}=\binom{2n}{n}$.