Example of nCr: let n=7 and r=3, 7C3=7×6×5/(1×2×3)=7!/(4!3!)=7!/((7-3)!3!).
So nCr=n!/((n-r)!r!) and nCr+1=n!/((n-r-1)!(r+1)!), by replacing r with r+1.
But (n-r)!=(n-r)(n-r-1)! and (r+1)!=(r+1)r!.
nCr+1/nCr=[n!/((n-r-1)!(r+1)r!)]/[n!/((n-r)(n-r-1)!r!)]=[1/(r+1)]/[1/(n-r)]=(n-r)/(r+1). (The strikethrough terms cancel out.)
Therefore nCr+1=(n-r)/(r+1)(nCr) QED