T TooF Member Joined Jan 6, 2004 Messages 46 Location NSW Aug 9, 2004 #1 How do you prove (n-1)Ck + (n-1)C(k-1) = nCk ???? It looks a bit messy, but I'm assuming that you'll be able to write it out properly on paper? Thanks! This is pretty urgent, exms tomorrow!
How do you prove (n-1)Ck + (n-1)C(k-1) = nCk ???? It looks a bit messy, but I'm assuming that you'll be able to write it out properly on paper? Thanks! This is pretty urgent, exms tomorrow!
nick1048 Mè çHöP ŸèW Joined Apr 29, 2004 Messages 1,614 Location The Mat®ix Ordinates: Sector 1-337- Statu Gender Male HSC 2005 Aug 9, 2004 #2 sub a different pro numeral for Ck, say x for instance and clean up the algebra...
T TooF Member Joined Jan 6, 2004 Messages 46 Location NSW Aug 9, 2004 #3 This isn't an algebra question though - its the binomial theorem you can't just stick in an x in place of Ck (or can you? :S)
This isn't an algebra question though - its the binomial theorem you can't just stick in an x in place of Ck (or can you? :S)
Supra secksy beast Joined Sep 27, 2003 Messages 2,399 Location On Top. Gender Male HSC 2004 Aug 9, 2004 #4 dont u just use the fact that nCk= n!/k!(n-k)! and manipul8 both sides
C coca cola Guest Aug 9, 2004 #5 you can find a simple proof in math in focus, page 424 if you got the new ed.
withoutaface Premium Member Joined Jul 14, 2004 Messages 15,098 Gender Male HSC 2004 Aug 9, 2004 #6 (n-1)!/[k!(n-1-k)!]+(n-1)!/[(k-1)!(n-k)!]=n!/[k!(n-k)!] n-k+k=n multiply both sides by (n-1)!/[k!(n-k)!] n=n, as required
(n-1)!/[k!(n-1-k)!]+(n-1)!/[(k-1)!(n-k)!]=n!/[k!(n-k)!] n-k+k=n multiply both sides by (n-1)!/[k!(n-k)!] n=n, as required