Equations with factorial numbers and combinatorial numbers

Let's now see equations that contain factorials or combinatorial numbers in general. There can be simple equations or especially difficult, therefore we have to be careful when they are generated. Let's see first how a simple equation is solved.

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

we must, then, solve the quadratic equation:



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

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 $$x = 9$$.

Let's see another slightly more complicated example. To resolve: $$$ \begin{pmatrix} x \\ 2 \end{pmatrix}+ \begin{pmatrix} x \\ 3 \end{pmatrix} =x+1$$$

We apply the Stifel formula to the first member: $$$ \begin{pmatrix} x \\ 2 \end{pmatrix}+ \begin{pmatrix} x \\ 3 \end{pmatrix} = \begin{pmatrix} x+1 \\ 3 \end{pmatrix} = x+1$$$

and we develop the combinatorial number: $$$\begin{pmatrix} x+1 \\ 3 \end{pmatrix} = \dfrac{(x+1)!}{3!(x-2)!} = \dfrac{(x+1)\cdot x \cdot (x-1)\cancel{(x-2)}}{3!\cancel{(x-2)!}} =x+1$$$

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

which we solve: $$$x=\dfrac{1\pm\sqrt{1+24}}{2}=\dfrac{1\pm5}{2} \Rightarrow x_1=3, \ x_2=-2$$$

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

As we were saying at the beginning, to construct an equation where factorials or combinatorial numbers are involved is a tricky subject. In this last example we are going to see how to do it.

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:

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