Question

prove that c (n, r)+ c (n, r_1)=c(n+1, r)

Answer 1

$nC_{r} + nC_{r-1} = \frac{n!}{(n-r)!r!} + \frac{n!}{(n-(r-1))!(r-1)!}$
[tex]= \frac{n!}{(n-r)!r!} + \frac{n!}{(n-r+1)!(r-1)!} [/tex]
$= \frac{n!}{(n-r)!r(r-1)!} + \frac{n!}{(n-r+1)(n-r+1-1)(r-1)!}$
[tex]= \frac{n!}{(n-r)!r(r-1)!} + \frac{n!}{(n-r+1)(n-r)!(r-1)!} \\ \\ = \frac{n!}{(n-r)!(r-1)!} \left \{ \frac{1}{r} + \frac{1}{(n-r+1)} \right \} \\ \\ = \frac{n!}{(n-r)!(r-1)!} \left \{ \frac{n-r+1-r}{r(n-r+1)} \right \} \\ \\ = \frac{n!}{(n-r)!(r-1)!} \left \{ \frac{(n+1)}{r(n-r+1)} \right \} \\ \\ = \frac{(n+1)n!}{(n-r+1)(n-r)!r(r-1)!} ---------1[/tex] 
[tex] (n+1)C_{r} = \frac{(n+1)!}{(n+1-r)!r!} \\ \\ = \frac{(n+1)(n+1-1)!}{(n+1-r)(n+1-r-1)!r(r-1)!} \\ \\ = \frac{(n+1)n!}{(n-r+1)(n-r)!r(r-1)!}--------2 [/tex]
[tex]1 = 2 \\ \\ nC_{r} + nC_{(r-1)} = (n+1)C_{r} [/tex]