The remainder when ∑k=010121012 k2024 k∑r=k2024r k2024 r is divided by 49 is (where rn=n!(n−r)!r!)

GE=∑k=01012(Ck 1012∑r=k2024Ck rCr 2024Ck 2024)=∑k=01012Ck 1012(1+1)2024−k =61012=(7−1)1012=49q+22

The remainder when ∑k=010121012 k2024 k∑r=k2024r k2024 r is divided by 49 is (where rn=n!(n−r)!r!)

GE=∑k=01012(Ck 1012∑r=k2024Ck rCr 2024Ck 2024)=∑k=01012Ck 1012(1+1)2024−k =61012=(7−1)1012=49q+22

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The remainder when $\sum_{k = 0}^{1012} (\frac{(\begin{matrix} 1012 \\ k \end{matrix})}{(\begin{matrix} 2024 \\ k \end{matrix})} \sum_{r = k}^{2024} ((\begin{matrix} r \\ k \end{matrix}) (\begin{matrix} 2024 \\ r \end{matrix})))$ is divided by 49 is
(where $(_{r}^{n})$ $= \frac{n!}{(n - r)! r!}$ )

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$G E = \sum_{k = 0}^{1012} ({}^{1012}C_{k} \sum_{r = k}^{2024} \frac{{}^{r}C_{k} {}^{2024}C_{r}}{{}^{2024}C_{k}}) = \sum_{k = 0}^{1012} {}^{1012}C_{k} {(1 + 1)}^{2024 - k} = 6^{1012} = {(7 - 1)}^{1012} = 49 q + 22$

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The remainder when ∑k=010121012 k2024 k∑r=k2024r k2024 r is divided by 49 is (where rn=n!(n−r)!r!)

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The remainder when $\sum_{k = 0}^{1012} (\frac{(\begin{matrix} 1012 \\ k \end{matrix})}{(\begin{matrix} 2024 \\ k \end{matrix})} \sum_{r = k}^{2024} ((\begin{matrix} r \\ k \end{matrix}) (\begin{matrix} 2024 \\ r \end{matrix})))$ is divided by 49 is
(where $(_{r}^{n})$ $= \frac{n!}{(n - r)! r!}$ )