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Let

$x = \frac{n!}{(n-r)!}$

If the value of x and r is known to me how do I find the minimum value of n that will satisfy the equation? I tried simplifying the equation by plugging in

$\frac{n!}{(n-r)!}\ =\ (n-r+1)\ *\ (n-r+2)\ ....\ *\ n$

This will result in a polynomial equation but since the value of r can be quite large the degree of this polynomial equation can be very high, how am I supposed to proceed from here?

Nakul Ram
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  • I guess you may try to find a prime factorization of factorials and $x$. And compare it. – openspace Jun 09 '18 at 15:44
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    You said the value of $x$ and $r$ are known and you want to find the minimum value of $r$. I think you mean either the value of $x$ and $n$ are known and you want $r$, or the value of $x$ and $r$ are known and you want $n$. – user1390 Jun 09 '18 at 15:59
  • Yeah sorry my bad! I need minimum value of n when n!/(n-r)! and r is known. – Nakul Ram Jun 09 '18 at 16:11
  • If $r\ne 0$ then $f(n)=n!/(n-r)!$ (for $\min(n,n-r)\geq 0$) is strictly increasing if $r>0 $ and strictly decreasing if $r<0$ and there is at most one $n$ for which $f(n)=x.$ – DanielWainfleet Jun 09 '18 at 16:38
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    See also https://math.stackexchange.com/questions/2345968/inverting-binomial-coefficients – lhf Jun 09 '18 at 18:00
  • for the record, $n$ is unique in that case, it can me minumum often different than a maximum if $x$ is given, $r,n$ to be guessed. – Abr001am Jun 10 '18 at 04:25

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