Question

Solve the following equation:

$C_3^{x+1}:C_4^x=6:5$, then $xϵN$

Answer 1

${}^{x+1}{C}_3{:}^x{C}_4=6:5⟹\frac{(x+1)!}{(x-2)!3!}\times \frac{4!(x-4)!}{x!}=\frac{6}{5}⟹\frac{24(x+1)}{6(x-2)(x-3)}=\frac{6}{5}$

$\frac{x+1}{(x-2)(x-3)}=\frac{3}{10}⟹3(x^2-5x+6)=10(x+1)⟹3x^2-25x+8=0⟹x=8,\frac{1}{3}$

$\because x\in N,x=8$