Question

The equation whose roots are the values of $r$ satisfying the equation ${}^{69}{C}_{3r-1}-{}^{69}{C}_{r^2}={}^{69}{C}_{r^2-1}-{}^{69}{C}_{3r}$ is

• A. $x^3-10x^2+21x=0$
• B. $x^2+10x+21=0$
• C. $x^2-10x+21=0$
• D. $x^2-21x+10=0$

Answer 1

The correct option is C $x^2-10x+21=0$

Given, ${}^{69}{C}_{3r-1}-{}^{69}{C}_{r^2}={}^{69}{C}_{r^2-1}-{}^{69}{C}_{3r}$
$\Rightarrow {}^{69}{C}_{3r}+{}^{69}{C}_{3r-1}={}^{69}{C}_{r^2}+{}^{69}{C}_{r^2-1}$
$\Rightarrow {}^{70}{C}_{3r}={}^{70}{C}_{r^2}(\because {}^n{C}_r+{}^n{C}_{r-1}={}^{n+1}{C}_r)$
$\Rightarrow 3r=r^2$ or $3r+r^2=70$
$\Rightarrow r(r-3)=0$ or $(r-7)(r+10)=0$
$\Rightarrow r=0,3$ or $r=7,-10$

But for $r=0,-10;{}^{69}{C}_{3r-1}$ is not defined.
$\therefore r=3,7$
Equation whose roots are $3,7$ is
$x^2-(3+7)x+3\cdot 7=0$
$\Rightarrow x^2-10x+21=0$