Factorial Problem

**venkateshm710** · August 19th 2011, 03:39 AM

Hi All,
I want to know the value of below:

Sum of [(n-2)! / (n-k)!], where n=59 and k begins with 2 and ends with 59.

**pickslides** · August 19th 2011, 03:49 AM

You mean $\displaystyle \sum_{k=2}^{59}\frac{(59-2)!}{(59-k)}$ ?

**venkateshm710** · August 19th 2011, 03:53 AM

Originally Posted by pickslides

You mean $\displaystyle \sum_{k=2}^{59}\frac{(59-2)!}{(59-k)}$ ?

yes absolutely !!

**venkateshm710** · August 19th 2011, 03:54 AM

(59-k) !

**venkateshm710** · August 19th 2011, 03:55 AM

Originally Posted by pickslides

You mean $\displaystyle \sum_{k=2}^{59}\frac{(59-2)!}{(59-k)!}$ ?

I want to know the value for above

**Plato** · August 19th 2011, 04:01 AM

Originally Posted by pickslides

You mean $\displaystyle \sum_{k=2}^{59}\frac{(59-2)!}{(59-k)}$ ?

Should the be $!$ in the denominator?
$\sum_{k=2}^{59}\frac{(59-2)!}{(59-k)\color{red}!}$

**venkateshm710** · August 19th 2011, 04:02 AM

yes .

**Plato** · August 19th 2011, 04:16 AM

Here is the value.

$\sum_{k=2}^{59}\frac{(59-2)!}{(59-k)!}=110163588853530184980727482893521304479399855 114493057608241578732074256106650$

**venkateshm710** · August 19th 2011, 05:48 AM

Originally Posted by Plato

Here is the value.

$\sum_{k=2}^{59}\frac{(59-2)!}{(59-k)!}=110163588853530184980727482893521304479399855 114493057608241578732074256106650$

Thank U man

**venkateshm710** · August 19th 2011, 05:49 AM

Hi How did u calculated this ??? can u tell me the procedure ??

Thread: Factorial Problem