Question

Determine $n$, if ${}^{2n}{C}_3{:}^n{C}_3=12:1$

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

Answer: (D)

$5$

${}^{2n}{C}_3{:}^n{C}_3=12:1$

first we calculate ${}^{2n}{C}_3$ and ${}^n{C}_3$ seperately

${}^{2n}{C}_3=\frac{2n!}{3!(2n-3)!}=\frac{2n(2n-1)(2n-2)(2n-3)!}{3\times 2\times 1\times (2n-3)!}$

$=\frac{2n(2n-1)(2n-2)}{6}$

${}^n{C}_3=\frac{n!}{3!(n-3)!}=\frac{n(n-1)(n-2)(n-3)!}{3\times 2\times 1\times (n-3)!}$

$=\frac{n(n-1)(n-2)}{6}$

${}^{2n}{C}_3{:}^n{C}_3=12:1$

$\therefore \frac{2n(2n-1)(2n-2)}{6}:\frac{n(n-1)(n-2)}{6}=12:1$

$\Rightarrow \frac{2n(2n-1)(2n-2)}{n(n-1)(n-2)}=12$

$\Rightarrow 2\times 2\frac{(2n-1)(n-1)}{(n-1)(n-2)}=12$

$\Rightarrow 8n-4=12n-24$

$\Rightarrow 4n=20$

$\Rightarrow n=\frac{20}{4}$

$\Rightarrow n=5$