Mathematics · 2019
JEE Main · 10 January 2019, Shift 2 · Q66
If Σ_r =0^25{^50 C_r ·^50- r C_25- r }= K (^50 C_25), then K is equal to:
If $\displaystyle \sum_{\mathrm{r}=0}^{25}\left\{{ }^{50} \mathrm{C}_{\mathrm{r}} \cdot{ }^{50-\mathrm{r}} \mathrm{C}_{25-\mathrm{r}}\right\}=\mathrm{K}\left({ }^{50} \mathrm{C}_{25}\right)$,
then K is equal to :
Official answer
(4)
$\displaystyle 2^{25}$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.