Question

Solve the following equation:
$C_3^x+C_4^x=11\cdot C_2^{x+1},xϵN$

Answer 1

$\frac{x!}{3!(x-3)!}+\frac{x!}{4!(x-4)!}=11\times \frac{(x+1)!}{2!(x-1)!}$

$\frac{x!}{3!(x-3)!}[\frac{1}{x-3}+\frac{1}{4}]=\frac{11}{2}x\times (x+1)$

$\frac{(x-3)(x-2)(x-1)x}{3}\left[\frac{4+x-3}{(x-3)\left(4\right)}\right]=11x(x+1)$

$\frac{(x-2)(x-1)}{3}\left[\frac{x+1}{4}\right]=11(x+1)$

$(x-2)(x-1)=11\times 12$

$x^2-3x+2=132$

$x^2-3x-130=0$

$x^2+10x-13x-130=0$

$x[x+10]-13[x+10]=0$

$(x-13)(x+10)=0$

$x=13$ or $x=-10$

But x cannot be negative so $x=13$