Problems from Equations with factorial numbers and combinatorial numbers

Development:

$$\begin{array}{rcl}{\displaystyle \frac{x!}{5!(x-5)!}}& =& 3{\displaystyle \frac{(x-1)!}{3!(x-4)!}}\\ {\displaystyle \frac{x\cancel{(x-1)!}}{5!\cancel{(x-5)!}}}& =& 3{\displaystyle \frac{\cancel{(x-1)!}}{3!(x-4)\cancel{(x-5)!}}}\\ {\displaystyle \frac{x}{5!}}& =& {\displaystyle \frac{3}{3!(x-4)}}\\ x(x-4)& =& {\displaystyle \frac{3\cdot 5!}{3!}}\\ x^2-4x& =& 60\end{array}$$

Then, we solve the equation: $$x={\displaystyle \frac{4\pm \sqrt{64+240}}{2}}={\displaystyle \frac{4\pm \sqrt{304}}{2}}$$

As the square root $\sqrt{304}=17.4356\dots$ 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.

Solution:

There is none.

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