I tried to simplify it with the second fraction by multiplying the top and bottom by "r" and i didn't work. Well at least I don't understand how to get there. Can you please just explain how...
n!r/(n-r-1)!(r-1)!r
goes to
n!r/(n-r+1)!r!
I get one put of it, that r x (r-1)! = r! but what about the (n-r-1)! -> (n-r+1)!
I checked over my previous post and the only thing I see wrong is that I forgot to put brackets in the denominator of the second fraction of the RS.
I had it on my paper but typing it out here one has to be sooooo careful.
so the RS should have said:
RS = n!/((n-r)!r!) + n!/((r-1)!(n-(r-1))!)
which is then
RS = n!/((n-r)!r!) + n!/((r-1)!(n-r+1)!)
I hope you can follow it now
yes i got now, thanks for the help.
To simplify the expression (n!r / (n-r-1)!(r-1)!r), you want to understand how (n-r-1)! can become (n-r+1)!. Let's break it down step by step:
1. Start with the expression (n!r / (n-r-1)!(r-1)!r).
2. Look at the term in the denominator (n-r-1)!. The (n-r-1) factorial can be expanded as follows:
(n-r-1)! = (n-(r+1))!
3. Notice that (n-(r+1))! simplifies to (n-r+1)!. This happens because when you subtract (r+1) from n, the result changes from (n-r-1) to (n-r+1).
4. Substitute the simplified term back into the original expression:
n!r / (n-r-1)!r = n!r / (n-(r+1))!r
= n!r / (n-r+1)!r
By making this substitution, you have transformed (n-r-1)! to (n-r+1)!, which simplifies the expression.