Verified Solution • 2025–26 Session moderate QJEEM24Mad9e3: JEE Main 2024 Mathematics - Class 11 JEEM24Mad9e3 Solution: _______. If 11C12+11C23+…+11C910=nm with gcd(n,m)=1, then n+m is equal to \text { If } \frac{{ }^{11} C_1}{2}+\frac{{ }^{11} C_2}{3}+\ldots+\frac{{ }^{11} C_9}{10}=\frac{n}{m} \text { with } \operatorname{gcd}(n, m)=1 \text {, then } n+m \text { is equal to } If 211C1+311C2+…+1011C9=mn with gcd(n,m)=1, then n+m is equal to _______. Step-by-Step Verified Solution amp;∑r=1911Crr+1amp;=112∑r=1912Cr+1amp;=112[212−26]=20356amp;∴m+n=2041\begin{aligned}& \sum_{\mathrm{r}=1}^9 \frac{{ }^{11} \mathrm{C}_{\mathrm{r}}}{\mathrm{r}+1} \\& =\frac{1}{12} \sum_{\mathrm{r}=1}^9{ }^{12} \mathrm{C}_{\mathrm{r}+1} \\& =\frac{1}{12}\left[2^{12}-26\right]=\frac{2035}{6} \\& \therefore \mathrm{m}+\mathrm{n}=2041\end{aligned}amp;r=1∑9r+111Cramp;=121r=1∑912Cr+1amp;=121[212−26]=62035amp;∴m+n=2041 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring