The remainder when $\sum _{k=0}^{1012}\left(\frac{\left(\begin{array}{c}1012\\ \\ k\end{array}\right)}{\left(\begin{array}{c}2024\\ \\ k\end{array}\right)}\sum _{r=k}^{2024}\left(\left(\begin{array}{c}r\\ \\ k\end{array}\right)\left(\begin{array}{c}2024\\ \\ r\end{array}\right)\right)\right)$  is divided by 49 is  
(where $\left(\underset{r}{\overset{n}{ }}\right)$$=\frac{n!}{(n-r)!r!}$)
 

$$\begin{gathered} GE=\sum _{k=0}^{1012}\left({}^{1012}C_k\sum _{r=k}^{2024}\frac{{}^{r}C_k{}^{2024}C_r}{{}^{2024}C_k}\right)=\sum _{k=0}^{1012}{}^{1012}C_k{(1+1)}^{2024-k} \\ =6^{1012}={(7-1)}^{1012}=49q+22 \end{gathered}$$