Question

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

• A. $2^{19}-\frac{{(}^{20}{C}_{10}{+}^{20}{C}_9)}{2}$
• B. $2^{19}-\frac{{(}^{20}{C}_{10}{+}^2\times 20C_9)}{2}$
• C. $2^{19}-\frac{{}^{20}{C}_{10}}{2}$
• D. $2^{20}-({20}_{C_{10}}+{20}_{C_9})$

Answer 1

The correct option is B $2^{19}-\frac{{(}^{20}{C}_{10}{+}^2\times 20C_9)}{2}$

${}^{20}{C}_0{+}^{20}{C}_1{+}^{20}{C}_2{+}^{20}{C}_4{+}^{20}{C}_{12}{+}^{20}{C}_{13}{+}^{20}{C}_{14}{+}^{20}{C}_{15}$

${\because }^n{C}_x{=}^n{C}_{n-x}$

${\because }^{20}{C}_{15}{=}^{20}{C}_5$

${}^{20}{C}_{14}{=}^{20}{C}_6$

${}^{20}{C}_{13}{=}^{20}{C}_7$

${}^{20}{C}_{12}{=}^{20}{C}_8$

$\therefore$ we get

${}^{20}{C}_0+C_1+C_2+C_4+C_5+C_6+C_7+C_8$

${\because }^{20}{C}_0+C_1+...C_8+C_9+C_{11}+..{.}^{20}{C}_{20}=2^{20}$

$2\{C_0+C_1+...C_9\}+C_{10}=2^{20}$

$2\{C_0+C_1+...C_8\}{+}^2{C}_9+C_{10}=2^{20}$

$\therefore C_0+C_1....C_8=\frac{2^{20}-{(}^2{C}_9+C_{10})}{2}$

$=2^{19}-\frac{{(}^{20}{C}_{10}+2^{20}{C}_9)}{2}$

Ans. (B)