Question

The value of ${}^{20}{C}_0+{}^{20}{C}_1+\dots +{}^{20}{C}_8$ 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. $2^{19}-\frac{(2\times {}^{20}{C}_{10}+{}^{20}{C}_9)}{2}$

Answer 1

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

Given
${}^{20}{C}_0+{}^{20}{C}_1+\dots +{}^{20}{C}_8$
We know that,
${}^{20}{C}_0+{}^{20}{C}_1+\dots +{}^{20}{C}_{20}={20}^{20}\Rightarrow 2[{}^{20}{C}_0+{}^{20}{C}_1+\dots {}^{20}{C}_9]+{}^{20}{C}_{10}=2^{20}\Rightarrow {}^{20}{C}_0+{}^{20}{C}_1+\dots {}^{20}{C}_9=\frac{2^{20}-{}^{20}{C}_{10}}{2}\Rightarrow {}^{20}{C}_0+{}^{20}{C}_1+\dots {}^{20}{C}_8=2^{19}-\frac{({}^{20}{C}_{10}+2{}^{20}{C}_9)}{2}$