Question

If ${}^{4n}{C}_{\text{2n}}{:}^{2n}{\text{C}}_{\text{n}}=\{1.3,.5,...(4\text{n}-1\left)\right\}:\{1,3,\mathrm{5....}(2\text{n}-1){\}}^{\lambda },then\lambda =$

• A. 1
• B. 2
• C. 3
• D. none of these

Answer 1

The correct option is B 2

${}^{4n}{C}_{2n}{:}^{2n}Cn=\{1,3,5,\dots .(4n-1)\};\{1,3,5,\dots .(2n-1)\}$

We have $\frac{2n!}{n!}=\frac{2n(2n-1)(2n-1)\dots .3.2.1}{n!}$

$=[\mathrm{2.4.6}\dots .(2n-2)\left(2n\right)][\mathrm{1.3.5}\dots .(2n-1)]/n!$

Writing the odd terms and even terms we get

$\frac{2n!}{n!}=2^n[\mathrm{1.2.3}\dots .(n-1).n][\mathrm{1.3.5}\dots .(2n-1)]/n!$

$=\frac{2^{n}n![\mathrm{1.3.5}\dots .(2n-1).n]}{n!}=[\mathrm{1.3.5}\dots .(2n-1)]$

Replacing $n$ by $2n$,

$\frac{4n!}{2n!}=2^{2n}[\mathrm{1.3.5}\dots .(4n-1)]$

Now we have ${}^{4n}{C}_{2n}{:}^{2n}{C}_n=\frac{4n!}{2n!2n!}\times \frac{n!n!}{2n!}=\frac{4n!}{2n!}.{\left(\frac{n!}{2n!}\right)}^2$

Now ${}^{4n}{C}_{2n}{!}^{2n}{C}_n=2^{2n}\frac{[\mathrm{1.3.5}\dots .(4n-1)]}{\left[2^n(\mathrm{1.3.5}\dots .(2n-1\left)\right)\right]}$

$=\frac{[\mathrm{1.3.5}\dots .(4n-1)]}{{[\mathrm{1.3.5}\dots .(2n-1)]}^2}\Rightarrow \lambda =2$