I need to understand the stages of this simplification

  • 0 votes

Thank you, I have copied exactly the solution of a part of a question from a CD:

nC2(2k)n-2=nC3(2k)n-3

    n!   

(n-2)!2!

(2k)n-2=

   n!     

(n-3)3!

(2k)n-3

(2k)n-2 = (n-2)!2!      (use laws of indices)

               (n-3)!3!

 

(2k)n-3

(2k)1 = (n-2)!2!      =[(n-2) x (n-3)!]

            (n-3)!3!

2k=

(n-2)x 2

    6

    3

3x2k=n-2

6n=n-2

N=6k+2

sorry the fractions did not show up :(

Posted Sat 2nd March, 2013 @ 13:57 by Cymbeline
Edited by Cymbeline on Sat 2nd March, 2013 @ 13:58