Question

The value of ${}^{20}{C}_0{+}^{20}{C}_1{+}^{20}{C}_2{+}^{20}{C}_3{+}^{20}{C}_4{+}^{20}{C}_{12}{+}^{20}{C}_{13}{+}^{20}{C}_{14}{+}^{20}{C}_{15}$ is

• A. $2^{19}-\frac{{(}^{20}{C}_{10}{+}^{20}{C}_9)}{2}$
• B. $2^{19}-\frac{{(}^{20}{C}_{10}+2{\times }^{20}{C}_9)}{2}$
• C. $2^{19}-\frac{{}^{20}{C}_{10}}{2}$
• D. none of these

Answer 1

The correct option is B $2^{19}-\frac{{(}^{20}{C}_{10}+2{\times }^{20}{C}_9)}{2}$

Consider $(1+x{)}^{20}$

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

Substituting $x=1$ we get

$2^{20}={}^{20}{C}_0+{}^{20}{C}_1+{}^{20}{C}_2...{}^{20}{C}_{20}$

By using the property of the binomial coefficient

${}^n{C}_r={}^n{C}_{n-r}$ we get

$2^{20}=2({}^{20}{C}_0+{}^{20}{C}_1+{}^{20}{C}_2+{}^{20}{C}_3+{}^{20}{C}_4+{}^{20}{C}_5+{}^{20}{C}_6+{}^{20}{C}_7+{}^{20}{C}_8+{}^{20}{C}_9)+{}^{20}{C}_{10}$

${}^{20}{C}_{10}$ is left out since it is the middle term.

Hence

$2^{20}-{}^{20}{C}_{10}=2({}^{20}{C}_0+{}^{20}{C}_1+{}^{20}{C}_2+{}^{20}{C}_3+{}^{20}{C}_4+{}^{20}{C}_5+{}^{20}{C}_6+{}^{20}{C}_7+{}^{20}{C}_8+{}^{20}{C}_9)$

$2^{19}-\frac{{}^{20}{C}_{10}}{2}={}^{20}{C}_0+{}^{20}{C}_1+{}^{20}{C}_2+{}^{20}{C}_3+{}^{20}{C}_4+{}^{20}{C}_5+{}^{20}{C}_6+{}^{20}{C}_7+{}^{20}{C}_8+{}^{20}{C}_9$

$2^{19}-\frac{{}^{20}{C}_{10}+2{}^{20}{C}_9}{2}={}^{20}{C}_0+{}^{20}{C}_1+{}^{20}{C}_2+{}^{20}{C}_3+{}^{20}{C}_4+{}^{20}{C}_5+{}^{20}{C}_6+{}^{20}{C}_7+{}^{20}{C}_8$

Again applying ${}^n{C}_r={}^n{C}_{n-r}$ we get

$2^{19}-\frac{{}^{20}{C}_{10}+2{}^{20}{C}_9}{2}={}^{20}{C}_0+{}^{20}{C}_1+{}^{20}{C}_2+{}^{20}{C}_3+{}^{20}{C}_4+{}^{20}{C}_{15}+{}^{20}{C}_{14}+{}^{20}{C}_{13}+{}^{20}{C}_{12}$