expand (a+b)!

frankbaugh · 2015-02-14 10:10:22

This is kind of basic compared to most of the stuff on the forum, but I was trying to find a definitive answer of n choose n-1. so n!/(n!(n-n-1)!) = n!/n!(-1)! and negatives don't have factorials. So I tried to expand (n-n-1)! To no avail. Does anyone have an answer? I tried (a+b)!(a+b-1)!(a+b-2)!.... (a+b-a+b)! But no idea how to do that.... Please help me!!

zetafunc · 2015-02-14 10:32:39

Your problem is simply dealing with the brackets (and it seems you made a small error in the formula).

i.e.

frankbaugh · 2015-02-14 10:36:12

Surely (n-(n-1))! = (n-n-1)! = (0-1)! = (-1)! With bidmas

zetafunc · 2015-02-14 10:53:03

No, you're not distributing the minus sign over the bracket.

frankbaugh · 2015-02-14 11:04:09

Ok but can you still expand (a+b)!

zetafunc · 2015-02-14 11:08:54

I can't do anything more for you without additional information about a,b. Is this what you were looking for?

frankbaugh · 2015-02-14 11:32:12

So is this as far as maths can possibly take us? What if (a+b)^(a+b)(-(1*2*3*4...a+b))

zetafunc · 2015-02-14 21:04:31

I don't understand what you're asking. Is this related to a problem you're working on? If so, it would help if you posted the problem in full detail, or provided a bit more information.

frankbaugh · 2015-02-15 07:17:58

For example, (a+b)^n = sigma bla bla bla. so what is (a+b)! Expanded?

bobbym · 2015-02-15 08:16:15

Hi;

If you mean a similar identity to ( a + b )^n I am not aware of one. There are Taylor series expansions for that though.

Math Is Fun Forum

#1 2015-02-14 10:10:22

expand (a+b)!

#2 2015-02-14 10:32:39

Re: expand (a+b)!

#3 2015-02-14 10:36:12

Re: expand (a+b)!

#4 2015-02-14 10:53:03

Re: expand (a+b)!

#5 2015-02-14 11:04:09

Re: expand (a+b)!

#6 2015-02-14 11:08:54

Re: expand (a+b)!

#7 2015-02-14 11:32:12

Re: expand (a+b)!

#8 2015-02-14 21:04:31

Re: expand (a+b)!

#9 2015-02-15 07:17:58

Re: expand (a+b)!

#10 2015-02-15 08:16:15

Re: expand (a+b)!

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