Hi,

Does anyone know whether it is possible to simplify a factorial as follows:

= (n-m)! / ((i-m)!(n-m-i-m)!)

=(n-m)!/((i-m)!(n-i)!)

=(n-m)!/(i-m+n-i)!

=(n-m)!/(n-m)!

=1

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- Sep 2nd 2008, 03:21 AMNumber Cruncher 20Factorials
Hi,

Does anyone know whether it is possible to simplify a factorial as follows:

= (n-m)! / ((i-m)!(n-m-i-m)!)

=(n-m)!/((i-m)!(n-i)!)

=(n-m)!/(i-m+n-i)!

=(n-m)!/(n-m)!

=1 - Sep 2nd 2008, 03:57 AMflyingsquirrel
- Sep 2nd 2008, 04:16 AMNumber Cruncher 20
no, what ive been asked is to derive a closed formed expression for

(a) s = http://www.mathhelpforum.com/math-he...488598ba-1.gif for n= 0,1,2,....

(b) t = http://www.mathhelpforum.com/math-he...27c930f6-1.gif for n=0,1,2, .. and m = 0,1,2,...n

ive done the first part and showed that its equal to 1 when n = 0 and 0 when n not equal to 0. for the second part ive been given a hint that

http://www.mathhelpforum.com/math-he...fb3cf705-1.gif for http://www.mathhelpforum.com/math-he...195e4ccc-1.gif.

and to use this and part a to find a closed form expression for (b). i get the part a and the hint but i just dont see how to go from there to solving part b. - Sep 2nd 2008, 04:57 AMflyingsquirrel
- Sep 2nd 2008, 05:43 AMNumber Cruncher 20
I havent done summations for a while. Have i changed the variables properly so that we get:

- Sep 2nd 2008, 05:55 AMflyingsquirrel
- Sep 2nd 2008, 06:06 AMNumber Cruncher 20
Thank you for your help its really appreciated.

- Sep 2nd 2008, 06:29 AMNumber Cruncher 20
One last thing, but should the sum read:

(-1)^(j+2m) or is it right as (-1)^(j+m) - Sep 2nd 2008, 06:39 AMflyingsquirrel
- Sep 2nd 2008, 06:47 AMNumber Cruncher 20
sorry my bad i forgot the restrictions on the range of n and m so i thought it had to be for it to be 1.

Thanks again - Sep 2nd 2008, 06:52 AMNumber Cruncher 20
what i meant is that you get

and when n=m

if m and n are odd then t = -1 and when m and n are even then t is 1

is that the right answer.