Task B-4.3. Find all solutions to the equation

$$
\frac{1}{\binom{4}{x}}-\frac{1}{\binom{5}{x}}=\frac{1}{\binom{6}{x}}
$$
Solution. From the definition of the binomial coefficient, we have

$$
\binom{4}{x}=\frac{4!}{x!(4-x)!},\binom{5}{x}=\frac{5!}{x!(5-x)!},\binom{6}{x}=\frac{6!}{x!(6-x)!}
$$

so the given equation becomes

$$
\frac{x!(4-x)!}{4!}-\frac{x!(5-x)!}{5!}=\frac{x!(6-x)!}{6!}, \text { with the condition } x \leq 4
$$

The equation

$$
\frac{(4-x)!}{4!}-\frac{(4-x)!(5-x)}{4!\cdot 5}=\frac{(4-x)!(5-x)(6-x)}{4!\cdot 5 \cdot 6}
$$

after simplification becomes

$$
1-\frac{5-x}{5}=\frac{(5-x)(6-x)}{5 \cdot 6} \text { or } x^{2}-17 x+30=0
$$

The solutions to this equation are 15 and 2, but due to the condition $x \leq 4$, the only solution is $x=2$.