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How can $\frac{(an)!}{n!}$ be expressed in terms of $a!$, $n!$, $a$, $n$ (and maybe Pochhammers)?

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What makes you think that it can? –  Cameron Buie May 15 '13 at 12:52
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It can't. Let $a=2$, $n=3$, then your number has a factor $5$, which none of $a!$, $n!$, $a$, or $n$ have. –  Gerry Myerson May 15 '13 at 12:52
    
Can't we use Pochhammer symbols? –  Soham Chowdhury May 15 '13 at 12:54
    
If all you want to do is avoid $(an)!$, you have $[(a-1)n]!{an\choose n}$ or (with Pochhammer symbols) $\frac{(an)_{an}}{n!}$. –  vadim123 May 15 '13 at 13:07
    
@vadim123 make that an answer! –  Soham Chowdhury May 15 '13 at 13:53

3 Answers 3

If you don't insist on being exact, you can use Stirling's approximation. $$\frac{(an)!}{n!}\approx \frac {(an)^{an}e^n}{n^ne^{an}}\sqrt{a}=a^{an}n^{an-1}e^{-(an-1)}\sqrt a$$ which will be very close-within about a factor of $1+\frac 1{12n}$

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So the accuracy increases with an increase in $n$? –  Soham Chowdhury May 15 '13 at 13:31
    
Yes, it does. And for $n=10$ you are already within $1\%$ This is very useful for large factorials. –  Ross Millikan May 15 '13 at 13:42
    
Nice, thanks - but it unfortunately doesn't meet my needs. –  Soham Chowdhury May 15 '13 at 13:52

By OP's request, my above comment was that since $m!=(m)_m$, we may write the desired expression as $$\frac{(an)_{an}}{a!}$$

Note: this solution is not the one @lhf commented on, below.

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This argument is only correct if you assume that "in terms of" means "multiplicatively". In principle, there might be an expression involving addition and then other primes might show up. –  lhf May 15 '13 at 12:54

For $n>m$, $$ \frac{n!}{m!} = P(n,n-m) = \text{the number of permutations of $n-m$ in a set of $n$}. $$ This is one context in which it doesn't make sense to define "permutation" as a bijection from a set to itself, but rather one should define it as an ordered list of length $n-m$ of members of a set of size $n$.

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