Question

Evaluate the following :

(i) ${}^{14}{C}_3$ (ii) ${}^{12}{C}_{10}$

(iii) ${}^{35}{C}_{35}$ (iv) ${}^{n+1}{C}_n$

Answer 1

(i) ${}^{14}{C}_3$
$\frac{14!}{3!(14-3!)}({\because }^n{C}_r=\frac{n!}{r!(n-r)!})=\frac{14!}{3!11!}=\frac{14\times 13\times 12\times 11!}{3\times 2\times 1\times 11!}=\frac{14\times 13\times 12}{6}=364$

(ii) ${}^{12}{C}_{10}$
$=\frac{12!}{10!(12-10)!}({\because }^n{C}_r=\frac{n!}{r!(n-r)!})=\frac{12\times 11\times 10!}{10!\times 2\times 1}=66$

(iii) ${}^{35}{C}_{35}$
$=\frac{35!}{35!(35-35)!}({\because }^n{C}_r=\frac{n!}{r!(n-r)!})=1$

(iv) (v) ${\sum }_{r=1}^5{}^5{C}_r$
$={}^5{C}_1+{}^5{C}_2+{}^5{C}_3+{}^5{C}_4+{}^5{C}_5$
$=\frac{5!}{1!4!}+\frac{5!}{2!3!}+\frac{5!}{3!2!}+\frac{5!}{4!1!}+\frac{5!}{5!0!}({\because }^n{C}_r=\frac{n!}{r!(n-r!)})=5+\frac{5\times 4}{2}+\frac{5\times 4}{2}+5+1=5+10+10+5+1=31$