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

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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 .

Log in to reply

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

Log in to reply

Add ">>" before the text.

Log in to reply

Use \boxed{}

Log in to reply

Log in to reply

@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

Log in to reply

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 .

Log in to reply

Can you please explain the second step , not getting

Log in to reply

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 .

Log in to reply

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

Log in to reply

this out.

CheckLog in to reply

Log in to reply