Why is $\prod_{k=n}^{2n-3}\left(2n-k\right) = \prod_{j=3}^{n}j$?

Question

I am trying to figure out these two equal expressions from my textbook: $$\prod_{k=n}^{2n-3}\left(2n-k\right) = \prod _ {j=3} ^ {n}j$$

I have checked them and know that they apply, but what are the logical steps that lead me to the equation?

Answer 1

Let's write out the terms and see if we can observe a pattern. $$(2n-n)(2n-(n+1))(2n-(n+2))(2n-(n+3))...(2n-(2n-3))$$ Which simplifies to $$n(n-1)(n-2)(n-3)...(3)$$ Which is equal to the desired product, since multiplication is commutative.

Answer 2

Hint: Using a simple change variable: $$j = 2n - k$$

Hence, its range is: $$ 2n - n = n \geq j \geq 2n-(2n-3) = 3$$

Answer 3

\begin{equation} \prod_{k=n}^{2n-3}\left(2n-k\right) = (2n - n)(2n - (n+1))\ldots (2n - (2n-3)) = (n)(n-1)(n-2) \ldots 3 = \prod_{j=3}^n j \end{equation} Another way is to do a change of variable, take $j = 2n- k$, then the extremities go from $j_1$ to $j_2$ where $j_1 = 2n - n = n$ and $j_2 = 2n - (2n - 3) = 3$