Simplifying ${n \choose k} 2^k (n-k)_k$, where $(n)_i$ is the falling factorial

Question

Is there a different way of writing $${n \choose k} 2^k (n-k)_k$$ where $(n)_i = n(n-1)(n-2)\cdots(n-i+1)$ is the falling factorial?

One source says $$\frac{n!}{k!(k-1)!}2^k$$ and another says it is actually $$\frac{n!}{k!(n-2k)!}2^k$$

Which one has the correct denominator? I guess my main problem is seeing what $(n-k)_k$ translates to.

Answer 1

Well your expression is equal to $\frac{n!}{k!(n-k)!}2^k\underbrace{(n-k)\cdot...\cdot(n-2k+1)}_{=(n-k)_k}$ so the first terms in the denominator cancel out and you get that this is equal to $\frac{n!}{k!(n-2k)!}$. But note that for this to work $k$ has to be smaller than $\frac{n+1}{2}$.

Does this help?

Answer 2

You said your main problem is "seeing", I hope this helps:

$$ \begin{aligned} (n-k)_k &= (n-k) \cdots (n-k-k+1) \\ &= \frac{\overbrace{(n-k) \cdots (n-k-k+1)(n-k-k)\cdots 2 \cdot 1}^{(n-k)!}}{\phantom{(n-k) \cdots (n-k-k+1)}\underbrace{(n-k-k)\cdots 2 \cdot 1}_{(n-2k)!}} \\ &= \frac{(n-k)!}{(n-2k)!} \end{aligned}$$

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$$ \require{cancel} \binom{n}{k} (n-k)_k = \frac{n!}{k!\cancel{(n-k)!}}\cdot \frac{\cancel{(n-k)!}}{(n-2k)!} = \frac{n!}{k!(n-2k)!} $$