Mathematics - Binomial Theorem Question with Solution | TestHub

Since ${}^{10}{\mathrm{C}}_0+{}^{11}{\mathrm{C}}_1+{}^{12}{\mathrm{C}}_2+{}^{13}{\mathrm{C}}_3+\dots \dots \dots +{}^{999}{\mathrm{C}}_{989}$\ $={}^{1000}{\mathrm{C}}_{989}={}^{1000}{\mathrm{C}}_{11}$ (Since, ${}^{10}{\mathrm{C}}_0={}^{11}{\mathrm{C}}_0$ and ${}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}+{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}-1}={}^{\mathrm{n}+1}{\mathrm{C}}_{\mathrm{r}}$ ) So, ${}^{12}{\mathrm{C}}_2+{}^{13}{\mathrm{C}}_3+{}^{14}{\mathrm{C}}_4+\dots \dots +{}^{999}{\mathrm{C}}_{989}={}^{1000}{\mathrm{C}}_{11}-12$