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Question

Mathematics Question on Combinations

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

Answer

(i) 2nC3nC3=121\frac{^{2n}C_3}{^nC_3 }= \frac{12}{1}

⇒(2n)!3!(2n−3)!×3!(n−3)!n!=121⇒\frac{\left(2n\right)!}{3!\left(2n-3\right)!}\times\frac{3!\left(n-3\right)!}{n!}=\frac{12}{1}

⇒(2n)(2n−1)(2n−2)(2n−3)!(2n−3)!×(n−3)!n(n−1)(n−2)(n−3)!=12⇒\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

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

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

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

**(ii) **2nC3nC3=111 \frac{^{2n}C_3}{^nC_3} = \frac{11}{1}

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

⇒(2n)(2n−1)(2n−2)(2n−3)!(2n−3)!×(n−3)!n(n−1)(n−2)(n−3)!=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

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

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

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