can someone please simplify this if it equals the RHS, the RHS is in a combination so you need to actually plug it in the combination formula

http://img93.imageshack.us/i/combinationsproof.jpg/

 

Doesn't look like an equality to me -- plug in n = 1 and k = 0 (if you don't like those numbers, try n = 2 and k = 1).

 

\(\frac{(n-k)![(n-k+1)(n-k+2)-k(n-k+1)+k^2]}{k!(n-2k+2)!}\)

is the closest I can come. Haven't checked it carefully.

 

Nope that didn't work either ...

Maybe ...

\(\frac{(n-k)![(n-k+1)(n-k+2)-k(n-2k+1)-k^2]}{k!(n-2k+2)!}\)

What's the point of the problem anyway?

 

because i needed to proove this identity and it was not equaling for me, just wanted to see if it would work for anyone else to see if I was making a math error