I was solving a problem in discrete math and i got the answer down to:
$$26^8 = (n-7)(n-7)!$$
I was wondering how you would solve this for n, if that's even possible. Or do you think I made a mistake getting to this point?
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Welcome to MSE. How did you get to that point? – José Carlos Santos Mar 18 '21 at 10:22
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Trying to find the length n of a string such that the expected number of occurrences of another string of length 8 found in the n-string is 1. The letters in the string are chosen at random – greenspartains Mar 18 '21 at 10:26
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Hint
Is it possible, if you are forcing $n$ to be a positive integer $\geq 7$?
$(26)^8$ is divisible by both $2$ and $13$, but is relatively prime to each of $3,5,7,11$. Can a positive integer $m$, be found such that $m!$ satisfies these constraints?
user2661923
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Making the problem more general, the problem is to find $n$ such that $$(n-a)\,(n-a)!=k$$ where $k$ is a large number.
Making the approximation $(n-a)\,(n-a)!\sim (n-a+1)!$, we just need the inverse of the factorial.
On the site, @gary proposed a superb approximation of the inverse factorial (have a look here). Applied to the problem, this will give $$n \sim a+ \frac{{\log \left( {\frac{k}{{\sqrt {2\pi } }}} \right)}}{{W\left( {\frac{1}{e}\log \left( {\frac{k}{{\sqrt {2\pi } }}} \right)} \right)}} - \frac{1}{2}$$ where $W(.)$ is Lambert function.
Using $a=6$ and $k=26^8$ this would give,as an approximation $n \sim 20.3242$ while the "exact" solution of your problem is $n=20.3521$
Claude Leibovici
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