Equations with factorial numbers and combinatorial numbers
Solve the equation: $$\begin{pmatrix} x \\ 5 \end{pmatrix}=3 \begin{pmatrix} x-1 \\ 3 \end{pmatrix}$$
$$\begin{array}{rcl} \dfrac{x!}{5!(x-5)!} &=& 3\dfrac{(x-1)!}{3!(x-4)!} \\ \dfrac{x\cancel{(x-1)!}}{5!\cancel{(x-5)!}} &=& 3\dfrac{\cancel{(x-1)!}}{3!(x-4)\cancel{(x-5)!}} \\ \dfrac{x}{5!}&=&\dfrac{3}{3!(x-4)} \\ x(x-4)&=&\dfrac{3\cdot5!}{3!} \\ x^2-4x &=& 60 \end{array}$$
Then, we solve the equation: $$ x=\dfrac{4\pm\sqrt{64+240}}{2}=\dfrac{4\pm\sqrt{304}}{2}$$
As the square root $\sqrt{304}=17.4356\ldots$ is not an integer, $x$ will not be an integer either and therefore it cannot be part of a combinatorial number, that is, the equation has no solution.
There is none.