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

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

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

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

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

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