Question 1143956
<pre>
C(n+1,r) = C(n,r-1) + C(n,r)

which is often written this way:

{{{(matrix(2,1,n+1,r))=(matrix(2,1,n,r-1))+(matrix(2,1,n,r))}}}

We'll use the facts that 

{{{(matrix(2,1,M,S))= M!/(S!(M-S)!)}}}, {{{M!=M*(M-1)!}}} and {{{(M+1)!=(M+1)M!}}}

We'll prove the right side equals the left side:

{{{(matrix(2,1,n,r-1))+(matrix(2,1,n,r))}}}

{{{n!^""/((r-1)!(n^""-(r-1))!)+n!^""/(r!^""(n-r)!)}}}{{{matrix(3,1,"",""="","")}}}

{{{n!^""/((r-1)!^""(n-r+1)!)+n!^""/(r!^""(n-r)!)}}}

Rewrite (n-r+1)! in the left denominator and r! in the right denominator:

{{{n!^""/((r-1)!^""(n-r+1)(n-r)!)+n!^""/(r(r-1)!^""(n-r)!)}}}

The LCD is r(r-1)!(n-r+1)(n-r)!

{{{(n!^""r+n!(n-r+1))/(r(r-1)!^""(n-r+1)(n-r)!)}}}

Factor n! out of the numerator and rewrite the denominator:

{{{(n!(r^""+(n-r+1)))/(r!^""(n-r+1)!)}}}

{{{(n!(r^""+n-r+1))/(r!^""(n-r+1)!)}}}

{{{(n!^""(n+1))/(r!^""(n-r+1)!)}}}

Rewrite the numerator as (n+1)! and the (n-r+1)! as (n+1-r)!

{{{(n+1)!/(r!^""(n+1-r)!)}}}

That's the same as the original left side:

{{{(matrix(2,1,n+1,r))}}}

Edwin</pre>