JEE Main 2026 — 21 January, Evening Shift — Question 24

$\quad \prod_{\mathrm{r}=0}^{12}\left(\frac{1}{{ }^{15} \mathrm{C}_{\mathrm{r}}}+\frac{1}{{ }^{15} \mathrm{C}_{\mathrm{r}+1}}\right)=\prod_{\mathrm{r}=0}^{12} \frac{\frac{16}{\mathrm{r}+1} \cdot{ }^{15} \mathrm{C}_{\mathrm{r}}}{{ }^{15} \mathrm{C}_{\mathrm{r}} \cdot{ }^{15} \mathrm{C}_{\mathrm{r}+1}}$

$=\prod_{r=0}^{12} \frac{16}{(r+1) \cdot \frac{15}{r+1} \cdot{ }^{14} C_{r}}=\prod_{r=0}^{12} \frac{\left(\frac{16}{15}\right)}{{ }^{14} C_{r}}$ $=\frac{\left(\frac{16}{15}\right)^{13}}{{ }^{14} \mathrm{C}_{0} \cdot{ }^{14} \mathrm{C}_{1} \ldots{ }^{14} \mathrm{C}_{12}}$

$\Rightarrow \alpha=\frac{16}{15}$

$\Rightarrow 30 \alpha=32$