Question

The value of $(\frac{30}{0})(\frac{30}{10})+(\frac{30}{1})(\frac{30}{11})+(\frac{30}{2})(\frac{30}{12})+...+(\frac{30}{20})(\frac{30}{30})$ is, where $(\frac{n}{r})=n_{C_r}$

• A. $(\frac{30}{10})$
• B. $(\frac{30}{15})$
• C. $(\frac{60}{30})$
• D. $(\frac{31}{10})$

Answer 1

The correct option is A $(\frac{30}{10})$

To find

${(}^{30}{C}_0\left){(}^{30}{C}_{10}\right)+{(}^{30}{C}_1\left){(}^{30}{C}_{11}\right)+{(}^{30}{C}_2\left){(}^{30}{C}_{12}\right)+...+{(}^{30}{C}_{20}\left){(}^{30}{C}_{30}\right)$

($\because$ difference of lower suffixes $=10$)

$\because (1+x{)}^{30}={}^{30}{C}_0+{}^{30}{C}_{1}x+{}^{30}{C}_2{x}^2+...+{}^{30}{C}_{20}{x}^{20}+...+{}^{30}{C}_{30}{x}^{30}$ ...(i)

$\Rightarrow (x-1{)}^{30}={}^{30}{C}_0{x}^{30}-{}^{30}{C}_1{x}^{29}+...+{}^{30}{C}_{10}{x}^{20}-{}^{30}{C}_{11}{x}^{19}+{}^{30}{C}_{12}{x}^{18}+...+{}^{30}{C}_{30}$ ...(ii)

Multiplying equation (i) and equation (ii) and comparing coefficient of $x^{20}$, the we get,

${(}^{30}{C}_0\left){(}^{30}{C}_{10}\right)-{(}^{30}{C}_1\left){(}^{30}{C}_{11}\right)+{(}^{30}{C}_2\left){(}^{30}{C}_{12}\right)+...+{(}^{30}{C}_{20}\left){(}^{30}{C}_{30}\right)={}^{30}{C}_{10}$

$[\because (1+x{)}^{30}(x-1{)}^{30}=(x^2-1{)}^{30}=(1-x^2{)}^{30}]$

$=(\frac{30}{10})$