We need \(S = \sum_{r=0}^{5}\frac{\binom{11}{2r}}{2r+2}\). Note \(\frac{\binom{11}{2r}}{2r+2} = \frac{1}{12}\binom{12}{2r+2}\) because \(\frac{\binom{11}{2r}}{2r+2}=\frac{11!}{(2r)!(11-2r)!}\cdot\frac{1}{2r+2}=\frac{1}{12}\cdot\frac{12!}{(2r+2)!(10-2r)!}=\frac{1}{12}\binom{12}{2r+2}\). So \(S=\frac{1}{12}\sum_{r=0}^{5}\binom{12}{2r+2}=\frac{1}{12}\left[\binom{12}{2}+\binom{12}{4}+\binom{12}{6}+\binom{12}{8}+\binom{12}{10}+\binom{12}{12}\right]\). The sum of even-indexed binomials of \(\binom{12}{k}\) is \(2^{11}=2048\), and removing \(\binom{12}{0}=1\) gives \(2048-1=2047\). Thus \(S=\frac{2047}{12}\). Here \(m=2047,n=12\) and \(\gcd(2047,12)=1\). So \(m-n=2047-12=2035\).