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We can know clearly from

$$(1+X)^n=\sum\limits_{i=0}^{n}C(n,i)X^n$$

that

$$ \sum\limits_{i=0}^{n}C(n,i)=2^n.$$

Whereas, I want to know if there are any researched results about permutations in the similar case, i.e., what can we know about

$$ \sum\limits_{i=0}^{n}P(n,i).$$

I’m really curious about that, but have found no answers elsewhere. Any help will be sincerely appreciated!

Community
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  • Have you tried some small numerical values? It's always good to show your work in your question so we at least know you've tried it yourself :) –  Mar 16 '18 at 12:09
  • @vrugtehagel Yes, I did...But none has proved useful. I’m also wondering if this is a well-researched problem; if so, I’ll be grateful if any reference is provided. $\ddot\smile$ –  Mar 16 '18 at 12:14
  • You should still show results like that in your question. Even if it's not useful to you, it might be useful to others, and again, it shows you've tried the problem yourself first. –  Mar 16 '18 at 12:16
  • In the first expression the sum should go from 0, shouldn't it? – zoli Mar 16 '18 at 12:17
  • @zoli They were misspelled ;) –  Mar 16 '18 at 12:20
  • $P(n,i)$ exactly being ? – G Cab Mar 16 '18 at 12:22
  • @GCab see permutations. Essentially, $P(n,k)=\frac{n!}{(n-k)!}$ –  Mar 16 '18 at 12:23
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    See here – Moya Mar 16 '18 at 12:24

1 Answers1

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$\sum\limits_{i=0}^{n}P(n,i) = \sum\limits_{i=0}^{n}\frac{n!}{(n-i)!}= \sum\limits_{i=0}^{n}\frac{n!}{i!}=n!\sum\limits_{i=0}^{n}\frac{1}{i!}=n!e - n!\sum\limits_{i=n+1}^{\infty}\frac{1}{i!} $

so

$n!e - \sum\limits_{i=0}^{n}P(n,i) = n!\sum\limits_{i=n+1}^{\infty}\frac{1}{i!} < \sum\limits_{i=1}^{\infty}\frac{1}{(n+1)^i}$

so

$n!e - \sum\limits_{i=0}^{n}P(n,i) < \frac{1}{n}$

Since $\sum\limits_{i=0}^{n}P(n,i)$ is an integer, we have

$\sum\limits_{i=0}^{n}P(n,i) = \lfloor {n!e} \rfloor $

gandalf61
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  • Thanks for you time! But...I’m rather confused about your $P(n,i)=\dfrac{n!}{i!}$ (๑•_•๑) –  Mar 16 '18 at 12:51
  • I have changed the lower limit to 0 to be consistent with the updated question, and added an extra step - the switch from n!/(n-i)! to n!/i! is by reversing the order of summation. – gandalf61 Mar 16 '18 at 13:05
  • Benny, actually the denominator is (n - i)!, but the sum of all permutations is the same. – bipll Mar 16 '18 at 13:15