Question

The value of $(11\cdot {}^{10}{P}_0-12\cdot {}^{11}{P}_1+13\cdot {}^{12}{P}_2-\cdots -20\cdot {}^{19}{P}_9)+({}^{12}{P}_2-{}^{13}{P}_3+{}^{14}{P}_4-\cdots +{}^{20}{P}_{10})$ is equal to

• A. 11
• B. 11.0
• C. 11.00

Answer 1

Answer: (A), (B), (C)

Let $S=(11\cdot {}^{10}{P}_0-12\cdot {}^{11}{P}_1+13\cdot {}^{12}{P}_2-\cdots -20\cdot {}^{19}{P}_9)+({}^{12}{P}_2-{}^{13}{P}_3+{}^{14}{P}_4-\cdots +{}^{20}{P}_{10})\cdots \left(1\right)$ and
$S_1=(11\cdot {}^{10}{P}_0-12\cdot {}^{11}{P}_1+13\cdot {}^{12}{P}_2-\cdots -20\cdot {}^{19}{P}_9)\cdots \left(2\right)$

Using: $n\cdot {}^{n-1}{P}_{r-1}={}^n{P}_r$,
Equation $\left(2\right)$ becomes
$\Rightarrow S_1={}^{11}{P}_1-{}^{12}{P}_2+{}^{13}{P}_3-{}^{14}{P}_4+\cdots -{}^{20}{P}_{10}$
putting the value of $S_1$ in $\left(1\right)$
$\Rightarrow S=({}^{11}{P}_1-{}^{12}{P}_2+{}^{13}{P}_3-{}^{14}{P}_4+\cdots -{}^{20}{P}_{10})+({}^{12}{P}_2-{}^{13}{P}_3+{}^{14}{P}_4-\cdots +{}^{20}{P}_{10})$
$\Rightarrow S={}^{11}{P}_1=11$