Question

Determine n if
(i)  ²ⁿC_3:ⁿC₃ = 12: 1
(ii)  ²ⁿC₃: ⁿC₃ = 11: 1

Answer 1

⁽ⁱ⁾  $\frac{^{2n}C_3}{^nC_3 }= \frac{12}{1}$

$⇒\frac{\left(2n\right)!}{3!\left(2n-3\right)!}\times\frac{3!\left(n-3\right)!}{n!}=\frac{12}{1}$

$⇒\frac{\left(2n\right)\left(2n-1\right)\left(2n-2\right)\left(2n-3\right)!}{\left(2n-3\right)!}\times\frac{\left(n-3\right)!}{n\left(n-1\right)\left(n-2\right)\left(n-3\right)!}=12$

$⇒\frac{2\left(2n-1\right)\left(2n-2\right)}{\left(n-1\right)\left(n-2\right)}=12$

$⇒\frac{4\left(2n-1\right)\left(n-1\right)}{\left(n-1\right)\left(n-2\right)}=12$

$⇒\frac{\left(2n-1\right)}{\left(n-2\right)}=3$
$⇒2n-1=3\left(n-2\right)$
$⇒2n-1=3n-6$
$⇒3n-2n=-1+6$
$⇒n=5$

(ii) $\frac{^{2n}C_3}{^nC_3} = \frac{11}{1}$

$⇒\frac{\left(2n\right)!}{3!\left(2n-3\right)!}\times\frac{3!\left(n-3\right)!}{n!}=11$

$⇒\frac{\left(2n\right)\left(2n-1\right)\left(2n-2\right)\left(2n-3\right)!}{\left(2n-3\right)!}\times\frac{\left(n-3\right)!}{n\left(n-1\right)\left(n-2\right)\left(n-3\right)!}=11$

$⇒\frac{2\left(2n-1\right)\left(n-1\right)}{\left(n-1\right)\left(n-2\right)}$

$⇒\frac{4\left(2n-1\right)\left(n-1\right)}{\left(n-1\right)\left(n-2\right)}=11$

$⇒\frac{4\left(2n-1\right)}{n-2}=11$
$⇒4\left(2n-1\right)=11\left(n-2\right)$
$⇒8n-4=11n-22$
$⇒11n-8n=-4+22$
$⇒3n=18$
$⇒n=6$