Turning product sequences into factorials I am trying to figure out the steps between these equal expressions in order to get a more general understanding of product sequences:
$$\prod_{k=0}^{n}\left(3n-k\right) + \prod_{k=n}^{2n-3}\left(2n-k\right) = \prod_{j=2n}^{3n}j + \prod_{j=3}^{n}j =\frac{(3n)!}{(2n-1)!}+\frac{n!}{2}$$
I know that $ n! :=\prod_{k=1}^{n}k$ but I can't figure out how that helps me understand the above equation.
edit: Thank you for the great help! Another thing I don't understand, is how I get from $\prod_{k=0}^{n}\left(3n-k\right) + \prod_{k=n}^{2n-3}\left(2n-k\right)$ to $\prod_{j=2n}^{3n}j + \prod_{j=3}^{n}j$. Any help with understanding this is much appreciated, I will try to figure it out myself while I wait for answers.
 A: Let's evaluate a few terms for $$\prod_{k=0}^n (3n-k)$$
when $k=0$, we are multipliying $3n$.
when $k=1$, we are multipliying $3n-1$.
when $k=2$, we are multipliying $3n-2$.
and so on. Each time, the individual term reduces by $1$.
when $k=n$, we are multiplying $3n-n=2n$
Hence we are just multiplying every term from $2n$ to $3n$.
$$\prod_{k=0}^n (3n-k)=\prod_{j=2n}^{3n}j=\frac{(3n)!}{(2n-1)!}$$
A: 
We obtain for integers $n\geq 2$:
\begin{align*}
\color{blue}{\prod_{k=0}^{n}}&\color{blue}{\left(3n-k\right) + \prod_{k=n}^{2n-3}\left(2n-k\right)}\\
&=\prod_{k=0}^n(2n+k)+\prod_{k=n}^{2n-3}(k+3-n)\tag{1}\\
&\,\,\color{blue}{=\prod_{k=2n}^{3n}k+\prod_{k=3}^{n}k}\tag{2}\\
&=\left(\prod_{k=1}^{3n}k\right)/\left(\prod_{k=1}^{2n-1}k\right)+\left(\prod_{k=1}^n k\right)/\left(\prod_{k=1}^2k\right)\tag{3}\\
&\,\,\color{blue}{=\frac{(3n)!}{(2n-1)!}+\frac{n!}{2}}
\end{align*}

Comment:

*

*In (1) we change in both products the order of multiplication: $k\to (n-k)$ and $k\to ((2n-3)-k+n)$.


*In (2) we shift the left product by $2n$ via $k\to2n+k$ and we shift the right product by $-n+3$ via $k\to k+n-3$.


*In (3) we complete the product to start with factors from $1$ and compensate this with the product in the denominator.
We start with $n\geq 2$, since the upper limit of the right product in (1) is negative in case $n=1$ invalidating the claim.
