How to solve this:

Find The value of :: \(\displaystyle{\sum _{ 0 }^{ n }{ { { ^{ n }{ P } } }_{ r } }} \).

here P(n,r) means permutation of n things taken r at a time .

I have not study Binaomial chapter in detaied Yet . But I need this sumition whlile solving an question of PNC .

PLease Help , I have no idea for such summitions , Please Explain , Thanks a lot

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TopNewestWe can actually evaluate the sum as \(\Gamma(n+1,1)\) , where \(\Gamma(a,x)\) is the Upper Incomplete Gamma function .

\[\begin{align} \sum_{r=0}^{n} \binom{n}{r} r! &= \sum_{r=0}^{n} \binom{n}{r} \int_{0}^{\infty} e^{-x} x^r \mathrm{d}x\\ &= \int_{0}^{\infty} \sum_{r=0}^{n} \binom{n}{r} e^{-x} x^r \mathrm{d}x\\ &= \int_{0}^{\infty} e^{-x} (1+x)^{n} \mathrm{d}x\\ &=\boxed{e \cdot \Gamma(n+1,1)}\end{align}\] .

This solution has been inspired from the genius Pratik Shastri's solution to this problem .

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Do you know how to add a question in a white box?

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Add ">>" before the text.

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Use \boxed{}

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@KARAN SHEKHAWAT himself .

Nope. Let's askLog in to reply

Ohh , so it is gamma summition . Can we use any other approach ? Since I didn't Know anything about gamma function , or can you please tell something about this gamma function ? I'am nwe to it . Thanks

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Well even I am new to Gamma function . If you study the topic from Wolfram Mathworld , you will get some understanding . It's actually quite easy , we are just exploiting the properties of Gamma Function .

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Can you please explain the second step , not getting

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Well in the second step I've just taken the Summation into the integral .

\[\begin{align} \int_{0}^{\infty} \sum_{r=0}^{n} \binom{n}{r} e^{-x} x^r \mathrm{d}x\\ \end{align}\] and we all know that \[ \sum_{r=0}^{n} \binom{n}{r} x^r = (1+x)^{n} \] . And after that we just use the definition of Incomplete Upper Gamma Function .

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howdid you take the summation into integral , how is this valid , can you explain in more detail pleaseLog in to reply

x[i.e.dx] and the summation is independant ofx[it's dependant onr]so I don't think that we should have a problem taking it into the integral .Log in to reply

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this out.

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