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Wolfram has the following identity:

$$n! = \frac{(n+m)!}{(n+1)_m}$$

I’m not sure I’m parsing it correctly — I am assuming the denominator is supposed to be a falling factorial. No qualifications are given on the values of m and n, but I don’t know how to evaluate the denominator if $m > n$. Even if I assume $m \leq n$, it just doesn’t appear to be true:

$$3! = 6 \neq \frac{(3+2)!}{(3+1)_2} = \frac{120}{12} = 10$$

What am I missing here?

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The notation $(k)_m$ is called the Pochhammer's symbol, defined as

$$ (k)_m = k \cdot(k+1) \cdot \dots \cdot (k+m-1) , (k)_0 = 1 $$

Hence we have

$$ \frac{(n+m)!}{(n+1)_m} = \frac{(n+m)\dots(n+1)n!}{(n+1)\dots(n+1+m-1)} = \frac{(n+m)\dots(n+1)}{(n+1)\dots(n+m)}n! = n! $$

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  • $\begingroup$ So sometimes falling factorial notation actually means rising factorial. That's probably the worst notational inconsistency encountered so far :p $\endgroup$ – Joseph Garvin Jun 10 '18 at 18:03

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