JEE Main 2025 Jan 24 Shift 2, Mathematics Q13: Properties Of Binomial Coefficient

The $(r+1{)}^{\text{th}}$ term of $(1+x{)}^{2n-1}$ has coefficient $\left(\genfrac{}{}{0ex}{}{2n-1}{r}\right)$. So:

$A=\left(\genfrac{}{}{0ex}{}{2n-1}{29}\right),B=\left(\genfrac{}{}{0ex}{}{2n-1}{11}\right)$

The condition $2A=5B$ gives:

$2\left(\genfrac{}{}{0ex}{}{2n-1}{29}\right)=5\left(\genfrac{}{}{0ex}{}{2n-1}{11}\right)$

Take the ratio:

$\frac{\left(\genfrac{}{}{0ex}{}{2n-1}{29}\right)}{\left(\genfrac{}{}{0ex}{}{2n-1}{11}\right)}=\frac{5}{2}$

Using $\left(\genfrac{}{}{0ex}{}{2n-1}{r}\right)=\frac{(2n-1)!}{r!(2n-1-r)!}$:

$\frac{11!(2n-12)!}{29!(2n-30)!}=\frac{5}{2}$

$\frac{(2n-12)(2n-13)\cdots (2n-29)}{29\cdot 28\cdots 12}=\frac{5}{2}$

Both numerator and denominator contain 18 consecutive integers. The right side simplifies cleanly when $2n-12=30$ (so the numerator becomes $30\cdot 29\cdots 13$, while the denominator is $29\cdot 28\cdots 12$):

$\frac{30\cdot 29\cdots 13}{29\cdot 28\cdots 12}=\frac{30}{12}=\frac{5}{2}✓$

So $2n-12=30\Rightarrow n=21$.

Correct option: (2)