Approach to prove the factorial on LHS to the RHS?

Question

I am brushing up my factorial solving skills, and I came across these two problems-

1) Prove that

$11\cdot 12\cdot 13\cdot ...\cdot 30=30!/10!$

which I solved as-

$=11\cdot 12\cdot 13\cdot ...\cdot 30$

$=(1\cdot 2\cdot 3...\cdot 10\cdot 11\cdot 12\cdot 13\cdot ...\cdot 30)/(1\cdot 2\cdot 3\cdot ...\cdot 10)$

$=30!/10!$

Am I correct here?

2) Prove that

$(2n)!/n!=2^n\cdot (2n-1)\cdot (2n-3)\cdot (2n-5)....\cdot 5\cdot 3 \cdot 1$

which I tried to solve as-

$=2n\cdot (2n-1)\cdot (2n-2)\cdot (2n-3)\cdot (2n-4)..../n!$

$=2n\cdot (2n)^n\cdot n!\cdot (1\cdot 3\cdot 5.....)/n!$ [Separating the odd and even]

$=2n\cdot (2n-1)\cdot (2n-3)\cdot (2n-5)...$ [Combining back the (2n) and odd values and canceling n!/n!]

This is as far as I could go with this problem, can anyone guide on how to achieve the LHS=RHS for this sum?

Answer 1

The proof for $1)$ is nice. Then at $2)$ you have

$$\frac{\color{red}(2n\color{red})!}{n!}=2n\cdot (2n-1)\cdot (2n-2)\cdot (2n-3)\cdot (2n-4)\cdot ...\cdot 2\cdot 1/n!$$

Now separating the odd and the even factors.

$$\underbrace{2n\cdot (2n-2)\cdot (2n-4)\cdot (2n-6)\cdot \ldots 4\cdot 2}_{\textrm{n factors}}\cdot (2n-1)\cdot (2n-3)\cdot (2n-5)\cdot...\cdot 3\cdot 1/n!$$

$$2^n\cdot \color{blue}{n\cdot (n-1)\cdot (n-2)\cdot (n-3)\cdot \ldots 2\cdot 1}\cdot (2n-1)\cdot (2n-3)\cdot (2n-5)\cdot...\cdot 3\cdot 1/\color{blue}{n!}$$

Cancelling out $\color{blue}{n!}$

$$2^n\cdot (2n-3)\cdot (2n-5)\cdot (2n-1)\cdot...\cdot 3\cdot 1$$

Answer 2

For the second one, and similar ones, I like to use the product notation.

$\begin{array}\\ \dfrac{(2n)!}{n!} &=\dfrac{\prod_{k=1}^{2n}k}{\prod_{k=1}^{n}k}\\ &=\dfrac{\prod_{k=1}^{n}(2k)\prod_{k=1}^{n}(2k-1)}{\prod_{k=1}^{n}k} \qquad\text{split into even and odd parts}\\ &=\dfrac{2^n\prod_{k=1}^{n}k\prod_{k=1}^{n}(2k-1)}{\prod_{k=1}^{n}k} \qquad\text{separate out 2}\\ &=2^n\prod_{k=1}^{n}(2k-1)\\ \end{array} $

Note that, since

$\begin{array}\\ \dfrac{(2n)!}{n!} &=\dfrac{\prod_{k=1}^{2n}k}{\prod_{k=1}^{n}k}\\ &=\prod_{k=n+1}^{2n}k\\ \end{array} $

we have $\prod_{k=n+1}^{2n}k =2^n\prod_{k=1}^{n}(2k-1) $.