MCQ
If $^{2n}{C_2}{:^n}{C_2} = 9:2$ and $^n{C_r} = 10$, then $r = $
- A$1$
- ✓$2$
- C$4$
- D$5$
$ \Rightarrow (2n)(2n - 1)2 = 9n(n - 1) \Rightarrow n = 5$
Now $^5{C_r} = 10 \Rightarrow r = 2$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$T=f(0)-f\left(\frac{\pi}{5}\right)+f\left(\frac{2 \pi}{5}\right)-f\left(\frac{3 \pi}{5}\right)+\ldots+f\left(\frac{8 \pi}{5}\right)-f\left(\frac{9 \pi}{5}\right) \text {. }$Then, $T$