Question

If ${}^{18}{C}_{15}+2\left({}^{18}{C}_{16}\right)+{}^{17}{C}_{16}+1={}^n{C}_3$ then $n$ is equal to

• A. 19
• B. 20
• C. 18
• D. 24

Answer 1

Answer: (B)

${}^{18}{}^{18}{C}_{15}+2\left({}^{18}{C}_{16}\right)+{}^{17}{C}_{16}+1={}^n{C}_3$
$\Rightarrow {}^{18}{C}_{15}+{}^{18}{C}_{16}+{}^{18}{C}_{16}+{}^{17}{C}_{16}+{}^{17}{C}_{17}={}^n{C}_3\left[\because {}^n{C}_n=1\right]$
$\Rightarrow {}^{18+1}{C}_{16}+{}^{18}{C}_{16}+{}^{17+1}{C}_{17}={}^n{C}_3\left[\because {}^n{C}_{r-1}+{}^n{C}_r={}^{n+1}{C}_r\right]$
$\Rightarrow {}^{19}{C}_{16}{}^{16}{}^{18}{C}_{16}{}^{16}+{}^{18}{C}_{17}={}^{17}{C}_3={}^{19}{C}_{16}+{}^{18+1}{C}_{17}={}^n{C}_3$
$\Rightarrow {}^{19}{C}_{16}+{}^{19}{C}_{17}={}^n{C}_3={}^{19+1}{C}_{17}={}^n{C}_3$
$\Rightarrow {}^{20}{C}_3={}^n{C}_3\left[\because {}^n{C}_{n-r}={}^n{C}_r\right]$
$\therefore n=20$