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Please refer to the attached image. I am not at all capable of understanding the solution given in this book for rth factorial moment about origin. Could someone please explain me how to get the third line from the second line in the given solution of book. Thank you in advance.

2 Answers 2

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Because of $x(x-1)\cdots(x-r+1)$ factor, the first $r$ terms of the sum are zero, so you are left with $\sum_{x=r}^n$...

Then:

$$\require{cancel}(x-1)\cdots(x-r+1){a\choose x}=x(x-1)\cdots(x-r+1)\frac{a(a-1)\cdots(a-x+1)}{x!}\\=\cancel{x(x-1)\cdots(x-r+1)}\frac{a(a-1)\cdots(a-r+1)(a-r)(a-r-1)\cdots(a-x+1)}{\cancel{x(x-1)\cdots(x-r+1)}(x-r)(x-r-1)\cdots1}\\ =a(a-1)\cdots(a-r+1)\frac{(a-r)(a-r-1)\cdots(a-x+1)}{(x-r)!}\\ =a(a-1)\cdots(a-r+1){a-r\choose x-r}$$

Lastly, $a(a-1)\cdots(a-r+1)$ and $a+b\choose n$ do not depend on $x$, and can be brought in the front of the sum.

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    Got it, need your help, now not able to get the product of combinations as from 3rd line to fourth line. – 2017-01-26
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    That is called [Vandermonde's identity](https://en.wikipedia.org/wiki/Vandermonde%27s_identity). Just plug in there $m=a-r, n=b, r=n-r, k=x-r$ – 2017-01-26
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I omit the terms which are the same in both lines.

Second line:

$x\cdot (x-1)\cdot \ldots \cdot (x-r+1)=\frac{x!}{(x-r)!} $

$^aC_x=\frac{a!}{x!(a-x)!}$

Product 1: $\frac{x!}{(x-r)!}\cdot \frac{a!}{x!(a-x)!}=\frac{1}{(x-r)!}\cdot \frac{a!}{(a-x)!}$

Third line:

$a\cdot (a-1)\cdot \ldots \cdot (a-r+1)=\frac{a!}{(a-r)!}$

$^{a-r}C_{x-r}=\frac{(a-r)!}{(x-r)!\cdot (a-x)!}$

Product 2: $\frac{a!}{(a-r)!}\cdot \frac{(a-r)!}{(x-r)!\cdot (a-x)!}=\frac{a!}{(x-r)!\cdot (a-x)!}$

Product 1 and product 2 are equal.

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    How to get the fourth line from third line. – 2017-01-26
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    @ShaileshSaxena I´m too late (again). But nevertheless I hope my answer has helped. – 2017-01-26
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    I need one more favour. Could you please tell me why they kept rth power of a in place of all the factors a (a-1).... (a-r+1) in last line ? – 2017-01-26
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    It is done for simplification. But this should be only an approximation. The equality sign is false here. – 2017-01-26