Milan solved the equation with binomial coefficients:
$${x \choose 2}+{x-1 \choose 2}=4$$
for $x\in\mathbb{N}$, as follows:
(1) First, according to the formula ${n \choose k}=\frac{n!}{k!⋅(n-k)!}$ , Milan converted the binomial coefficients in the equation into fractions.
$$\frac{x!}{2!\cdot(x-2)!}+\frac{(x-1)!}{2!\cdot(x-3)!}=4$$
(2) In both fractions, he factored out the factorial terms to be able to cancel them out later. $$\frac{(x-1)\cdot(x-2)!}{2!\cdot(x-2)!}+\frac{(x-2)\cdot(x-3)!}{2!\cdot(x-3)!}=4$$
(3) He canceled out the common terms in the first fraction using $(x-2)!$ and in the second fraction using $(x-3)!$.
$$\frac{x-1}{2!}+\frac{x-2}{2!}=4$$
(4) He multiplied both sides of the equation by $2! = 2$.
$$(x-1)+(x-2)=8$$
(5) Milan removed the parentheses and simplified the equation to its standard form. $$\begin{aligned} 2x -3&=8\cr 2x&=11\cr x&=5.5 \end{aligned}$$
(6) The solution to the equation is the number $5.5$, which is not an integer, $5.5\notin\mathbb{N}$. Therefore, $K=\emptyset$.
Did Milan make a mistake during the calculation? If yes, determine where.
Milan did not make a mistake and solved the problem correctly.
Milan made the first mistake when factoring the factorial expressions in step (2). It holds:
$$\begin{aligned} x!&=x\cdot(x-1)\cdot(x-2)!\cr (x-1)!&=(x-1)\cdot(x-2)\cdot(x-3)! \end{aligned}$$
Milan made the first mistake when simplifying the fractions in step (3). The equation should have been modified as follows:
$$\frac{(x-1)\cdot(x-2)}{2!}+\frac{(x-2)\cdot(x-3)}{2!}=4$$
Milan made the first mistake in step (4). It holds $2!=4$, so the equation should have been modified as follows: $$(x-1)+(x-2)=16$$
Decimal numbers can appear in binomial coefficients, and the solution to the equation is the found number $5.5$. Milan made a mistake only in the last step (6).
(1) First, according to the formula ${n \choose k}=\frac{n!}{k!\cdot(n-k)!}$ we convert the binomial coefficients in the equation into fractions.
$$\frac{x!}{2!\cdot(x-2)!}+\frac{(x-1)!}{2!\cdot(x-3)!}=4$$
(2) In both fractions, we factor out the factorial terms to be able to cancel them out later.
$$\frac{x\cdot(x-1)\cdot(x-2)!}{2!\cdot(x-2)!}+\frac{(x-1)\cdot(x-2)\cdot(x-3)!}{2!\cdot(x-3)!}=4$$
(3) We cancel out the common terms in the first fraction using $(x-2)!$ and in the second fraction using $(x-3)!$.
$$\frac{x\cdot(x-1)}{2!}+\frac{(x-1)\cdot(x-2)}{2!}=4$$
(4) We multiply both sides of the equation by $2! = 2$.
$$x\cdot(x-1)+(x-1)\cdot(x-2)=8$$
(5) Then we remove the parentheses and simplify the quadratic equation to its standard form. $$\begin{aligned} x^2-x+x^2-3x+2&=8\cr 2x^2-4x-6&=0\cr x^2-2x-3&=0\cr \end{aligned}$$
(6) We solve the quadratic equation.
$$D=b^2-4ac=4+12=16$$ $$n_1,\ n_2=\frac{-b\pm\sqrt{D}}{2a}=\frac{2\pm4}{2}$$ $$n_1=3,\ n_2=-1$$
(7) For a binomial coefficient ${n \choose k}$ to be valid:
- $n\in\mathbb{N}$,
- $k\in\mathbb{N}\cup\left{0\right}$,
- $n\geq k$.
in our case:
- $x\in\mathbb{N}$,
- $\left.\begin{aligned} &x\geq2\cr &x-1\geq2\Rightarrow x\geq3)\end{aligned}\right} x\geq3$
Therefore, we exclude the number $-1$ from the solution set. $$K=\left{3\right}$$