# Prove that ${2n \choose n}= 2{2n-1 \choose n}$

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!

• Hint: $(2n)!=2n\times (2n-1)!$ – J. W. Tanner Mar 15 at 20:10
• Notice that $$\frac{(2n - 1)!}{(n)! (n - 1)!} = \frac{(2n)!}{(2n) \cdot (n)! (n - 1)!}$$ – Viktor Glombik Mar 15 at 20:10
• @J.W.Tanner So I'm really new to this, can you explain how that's equivalent? I'm still learning how to work with factorials, and I'm at a real beginner's level. – Brownie Mar 15 at 20:20
• I put an explanation in my answer – J. W. Tanner Mar 15 at 20:34

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 $$2\binom{2n-1}{n}=\frac{(2n)!}{(n!)^2}=\binom{2n}{n}$$.

• This is great thanks for the explanation! I know this is a really basic question, but I'm kinda iffy on working with factorials. Could you explain how the multiplying and diving by 2n works? – Brownie Mar 15 at 20:19
• Well, if you multiply and divide by the same number you don't change the expression. Now let's see what we get from this. In the numerator we have $(2n-1)!\times (2n)=(2n)!$. In the denominator it is $(n)!(n-1)!\times (2n)=2(n)!(n-1)!n=2(n)!(n)!$. – Mark Mar 15 at 20:20
• oh, I see. So I didn't know / understand that (n-1)! * n just equals n!. Thanks for breaking it down! – Brownie Mar 15 at 20:25
• It follows from the definitions. $(n-1)!\times n=(1\times 2\times 3\times...\times (n-1))\times n=1\times 2\times...\times n=n!$. – Mark Mar 15 at 20:30

$$\text{We know from Pascal's triangle that }\binom m k=\binom{m−1}{k−1}+\binom{m−1}{k}.$$ $$\therefore\binom{2n}n=\binom{2n−1}{n−1}+\binom{2n−1}n=2\binom{2n−1}n$$

$$\text{since }\binom{2n-1}{n-1}=\binom{2n-1}n\text{ since } (2n-1)-(n-1)=n.$$

• Inspired by @Bernard – J. W. Tanner Mar 15 at 21:32

With very few calculations:

We can use the recurrence relation $$\binom nk=\frac nk\binom{n-1}{k-1},$$ which is the basis of the proof of the formula with factorials. In particular, $$\binom{2n}n= 2\binom{2n-1}{n-1}=2\binom{2n-1}n,\quad\text {since}\quad n=(2n-1)- (n-1).$$

• Quite a fine proof too! – Bernard Mar 15 at 20:59
• Ok, I put it as another answer. – J. W. Tanner Mar 15 at 21:09

Noting $$2n-n=n,$$ we have $${2n \choose n}=\frac{\color{red}{(2n)!}}{\color{green}{(n)!}(n)!}.$$

Also, noting $$(2n-1)-n=n-1,$$ we have $${2n-1 \choose n}=\frac{(2n-1)!}{(n)!(n-1)!}.$$

Finally, note that $$\color{red}{(2n)!}=2n\times(2n-1)!$$ and $$\color{green}{n!}=n\times(n-1)!$$ and $$\frac {2n}n=2$$ and we're done.

The last note is because $$m! = m\times(m-1)\times(m-2)\times...\times2\times1$$ and

$$(m-1)!=(m-1)\times (m-2)\times...\times2\times1$$,

so $$m!=m\times(m-1)!$$.