Sum of permutations!

I just found that sum of $n$ permutations with $i$ equals $e\Gamma(n+1, 1)$ . In other words,

$P(n,0) + P(n,1) + P(n, 2) + .... P(n,n) = e\Gamma(n+1, 1)$ . [where $\Gamma(x,y)$ is the incomplete gamma function].

Can anyone give a proof of this?

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
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Comments

Hi Kartik Sharma , see this or the solution to this question .

But if you aren't familiar with Gamma function , see the link I provided or see it here .

Hope I was useful !!!

A Former Brilliant Member - 6 years, 4 months ago

Thanks, that was helpful. You are quite good, solves almost all the problems.

Kartik Sharma - 6 years, 4 months ago

You are welcome.

A Former Brilliant Member - 6 years, 4 months ago

@A Former Brilliant Member – Then, can you help me here too? *If only you have time.

Kartik Sharma - 6 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$