# Question #e5721

Feb 18, 2015

Hello,

I suppose (x+1)C(3) means $\setminus {\textrm{C}}_{x + 1}^{3}$. So, your equation becomes

$\setminus \frac{\left(x + 1\right) x \left(x - 1\right)}{6} = \setminus \frac{x \left(x - 1\right)}{2}$

Obvious solutions are $x = 0$ and $x = 1$.

If $x \setminus \ne 0$ and $x \setminus \ne 1$, you can divide by $x \left(x - 1\right)$ because it's not zero, and you get

$\setminus \frac{x + 1}{6} = \setminus \frac{1}{2}$ therefore $x + 1 = 3$, and the last solution is $x = 2$.

Conclusion. The solutions are 0, 1 and 2.