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I'm studying for a test and in one of the solutions to a problem, Stirling's Approximation is used and allows the equation

$$ \frac{(q+N-1)!} {q! \cdot (N-1)!} $$

to become

$$ \frac{1}{q!} \cdot \frac{N}{N!} \cdot \frac{(q+N)!}{q+N} $$

I don't understand how the last two terms came to be. Is it a factorial identity or is there underlying math that hasn't been shown in the solution. Any help would be great.

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  • $\begingroup$ For $N=1,2,\dots$ (so not zero), $\frac{N!}{N}=\frac{N(N-1)\dots(2)(1)}{N}=(N-1)(N-2)\dots,(2)(1)=(N-1)!$. $\endgroup$ – Ian Sep 26 '16 at 0:35
  • $\begingroup$ It's quite an ordinary manipulation, if you know the definition of factorial. Just expand out the factorials, that will help you understand the reasoning of this transformation. $\endgroup$ – Teresa Lisbon Sep 26 '16 at 0:36
  • $\begingroup$ Your last term it seems to me should be $\frac{(q+N)!}{q+N}$ instead. If that's the case (and $q$ is a nonnegative integer) then there is no approximation going on here at all. $\endgroup$ – Ian Sep 26 '16 at 0:36
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$$\require{cancel}n!=\overbrace{1\times2\times3\times\dots\times(n-1)\times n}^n$$

$$\frac{n!}n=\frac{\overbrace{\color{green}{1\times2\times3\times\dots\times(n-1)}\times\color{red}{\cancel n}}^n}{\color{red}{\cancel n}}$$

$$(n-1)!=\overbrace{\color{green}{1\times2\times3\times\dots\times(n-1)}}^{n-1}$$

We also have:

$$\frac{n!}{p!}=\frac{\overbrace{\color{red}{\cancel{1\times2\times3\times\dots\times p}}\times\color{green}{(p+1)\times\dots\times(n-1)\times n}}^n}{\color{red}{\cancel{1\times2\times3\times\dots\times p}}}$$

$$=\color{green}{(p+1)\times\dots\times(n-1)\times n}$$

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  • $\begingroup$ I understand the answer, but could you please add some words on how did you get the last expression? $\endgroup$ – Ana S. H. Sep 26 '16 at 0:41
  • $\begingroup$ @Ana You cancel the $n$'s, and then what's left? I tried editing to make it more clear for you. $\endgroup$ – Simply Beautiful Art Sep 26 '16 at 0:41
  • $\begingroup$ :( I wonder why someone downvoted me. $\endgroup$ – Simply Beautiful Art Sep 26 '16 at 0:43
  • $\begingroup$ @Ana Did that answer your full concerns? $\endgroup$ – Simply Beautiful Art Sep 26 '16 at 0:49
  • $\begingroup$ I just wanted you to write that the second and third equations imply that $n!=n(n-1)!$. But it's OK. $\endgroup$ – Ana S. H. Sep 26 '16 at 2:21
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The factorial can resursively defined as $0! = 1$ and $$ n! = n\cdot(n-1)!$$ for $n\geq 1$.

An example is $4! = 4\cdot 3! = ... = 4\cdot 3\cdot 2\cdot 1.$ Hence by isolating the $(n-1)!$ term we get $$(n-1)! = \frac{n!}{n}.$$

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$$ \frac{(q+N-1)!} {q! \cdot (N-1)!}=\frac{1}{q!}\cdot\frac{1}{(N-1)!}\cdot(q+N-1)!= \\ =\frac{1}{q!}\cdot\frac{N}{N\cdot(N-1)!}\cdot\frac{(q+N-1)!\cdot(q+N)}{(q+N)}= \frac{1}{q!}\cdot\frac{N}{N!}\cdot\frac{(q+N)!}{(q+N)} $$

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