$\frac{C^{12}_n}{(n-1)n}=\frac{n!}{12!.(n-12)!.n.(n-1)}=\frac{(n-2)!}{12!.(n-12)!}=\frac{1}{12!}(n-2)...(n-11)$
=>[TEX]12!.S=1.2...10 + 2.3...11+...+2003.2004...2012[/TEX]
=>[TEX]11.12!.S=1.2...11+2.3...11.11+...+2003.2004...2012.11[/TEX]
[TEX]=1.2...11+2.3....11.(12-1) + 3.4...12.(13-2)+...+2003.2004...2012.(2013-2002)[/TEX]
[TEX]=1.2...11-1.2....11+2.3...12-2.3...12+3.4...13+...-2002.2003...2012+2003.2004...2013=2003.2004...2013[/TEX]
=>S=$\frac{2003.2004...2013}{11.12!}$