Equations with factorial numbers and combinatorial numbers

Example

$$x!=72\cdot (x-2)!$$ We move the $x$ to the first member of the equation and develop the factorials. $$\begin{array}{rcl}{\displaystyle \frac{x!}{(x-2)!}}& =& 72\\ {\displaystyle \frac{x(x-1)(x-2)!}{(x-2)!}}& =& 72\\ x(x-1)=x^2-x& =& 72\end{array}$$

we must, then, solve the quadratic equation:

$x^2-x-72=0$

$x={\displaystyle \frac{1\pm \sqrt{1+4\cdot 72}}{2}}={\displaystyle \frac{1\pm 17}{2}}$

$x_1=9$, $x_2=-8$

We reject the negative solution, since it does not make sense to talk about the factorial of a negative number. With that, the expected solution will then be $x=9$.

Example

Let's see another slightly more complicated example. To resolve: $$\left(\begin{array}{c}x\\ 2\end{array}\right)+\left(\begin{array}{c}x\\ 3\end{array}\right)=x+1$$

We apply the Stifel formula to the first member: $$\left(\begin{array}{c}x\\ 2\end{array}\right)+\left(\begin{array}{c}x\\ 3\end{array}\right)=\left(\begin{array}{c}x+1\\ 3\end{array}\right)=x+1$$

and we develop the combinatorial number: $$\left(\begin{array}{c}x+1\\ 3\end{array}\right)={\displaystyle \frac{(x+1)!}{3!(x-2)!}}={\displaystyle \frac{(x+1)\cdot x\cdot (x-1)\cancel{(x-2)}}{3!\cancel{(x-2)!}}}=x+1$$

Simplifying $(x+1)$ in both members the quadratic equation remains: $$x^2-x=-6\Rightarrow x^2-x-6=0$$

which we solve: $$x={\displaystyle \frac{1\pm \sqrt{1+24}}{2}}={\displaystyle \frac{1\pm 5}{2}}\Rightarrow x_1=3,x_2=-2$$

As we did before, we reject the negative solution because it is meaningless, and so the answer is $x=3$.

Example

We begin by deciding what the solution to the equation is going to be. For example $x=1$. The most simple equation possible with this solution is $x-1=0$. We square both members and we introduce new elements:

$$\begin{array}{rcl}(x-1{)}^2& =& 0\\ x^2-2x+1& =& 0\\ x(x-2)+1& =& 0\\ x(x-2)& =& -1\\ {\displaystyle \frac{x!}{(x-1)!}}\cdot {\displaystyle \frac{(x-2)!}{(x-3)!}}& =& -1\\ x!(x-2)!& =& -(x-1)!(x-2)!\end{array}$$