Combinatorial Proof - Equating binomials

Question

The question I am working on is:

Let $n, k \in \mathbb{Z}^+$ and $k < n$. Prove that:

$$\binom{n-1}{k-1} \binom{n}{k+1} \binom{n+1}{k} =\binom{n-1}{k} \binom{n}{k-1} \binom{n+1}{k+1}$$

I'm unsure of how to start this proof. My initial thought was to expand each binomial into its factorial form $$\frac{n!}{k!(n-k)!}$$ and see if I could equate them that way. It got quite confusing and messy though, and I stopped as I was wondering if there was an easier way to prove this or if my route was even correct. Any help is appreciated.

Answer 1

You had a good start. Expanding the LHS and rearranging:

$$\text{LHS}=\dfrac{(n-1)!}{(k-1)!(n-k)!}\dfrac{n!}{(k+1)!(n-k-1)!} \dfrac{(n+1)!}{k!(n+1-k)!} = \dfrac{(n-1)!}{k!(n-1-k)!}\dfrac{n!}{(k-1)!(n-(k-1))!}\dfrac{(n+1)!}{(k+1)!(n-k)!} = \text{RHS}$$

Answer 2

${(n-1)! \over (k-1)! (n-1-(k-1)!)}{n! \over (k+1)!(n-(k+1)!)}{(n+1)! \over k! (n+1-k)!}$

Reassociate:

${(n-1)! \over k! (n-k)!}{n! \over (k-1)!(n-(k-1))!}{(n+1)! \over (k+1)!(n+1-(k+1))!}$