Question

${}^{20}C_4+2.{}^{20}C_3+{}^{20}C_2-{}^{22}C_{18}=$

• A. $0$
• B. $1242$
• C. $7315$
• D. $6345$

Answer 1

The correct option is A

$0$

Explanation for the correct option:

Solve the given series using standard formulas

${}^{20}C_4+2.{}^{20}C_3+{}^{20}C_2–{}^{22}C_{18}$

$=\left[{}^{20}C_4+{}^{20}C_3\right]+\left[{}^{20}C_3+{}^{20}C_2\right]–{}^{22}C_{18}$

$=\left[{}^{21}C_4+{}^{21}C_3\right]–{}^{22}C_{18}$ $\left[\because {}^{\mathbf{n}}\mathit{C}_{\mathbf{r}}\mathbf{}\mathbf{+}\mathbf{}{}^{\mathbf{n}}\mathit{C}_{\mathbf{r}\mathbf{+}\mathbf{1}}\mathbf{}\mathbf{=}\mathbf{}{}^{\mathbf{n}\mathbf{+}\mathbf{1}}\mathit{C}_{\mathbf{r}\mathbf{+}\mathbf{1}}\right]$

$={}^{22}C_4–{}^{22}C_{18}={}^{22}C_{18}–{}^{22}C_{18}=\mathbf{0}$ $\left[\mathbf{\because }\mathbf{}{}^{\mathbf{n}}\mathit{C}_{\mathbf{r}}\mathbf{}\mathbf{=}\mathbf{}{}^{\mathbf{n}}\mathit{C}_{\mathbf{n}\mathbf{}\mathbf{–}\mathbf{r}}\right]$

Hence, Option ‘A’ is Correct.