# The value of sigma r=1 to 10 r*nCr/nCr-1

$\\\sum_{r=1}^{10}r{_{r}^{n}\textrm{C} \over _{r-1}^{n}\textrm{C}} \\=\sum_{r=1}^{10}(n-r+1) \\=10(n+1)-55 \\=10n-46 \\=5(2n-9)$