## Task 24/72

Determine all solutions in the domain of natural numbers for the system of equations

$$
\begin{aligned}
\binom{y+1}{x+1}-\binom{y}{x+1} & =6 \\
\binom{x}{x-2} \cdot \frac{2}{x-1}+\binom{y}{y-1} \cdot \frac{1}{y} & =3
\end{aligned}
$$
Due to $\binom{n}{n-k}=\binom{n}{k}$, the second of the given equations can be written in the form

$$
\binom{x}{2} \cdot \frac{2}{x-1}+\binom{y}{1} \cdot \frac{1}{y}=3
$$

from which, with

$$
\binom{x}{2}=\frac{x(x-1)}{2} \quad ; \quad\binom{y}{1}=y
$$

it follows that

$$
\frac{2 x(x-1)}{2(x-1)}+\frac{x}{y}=3
$$

Obviously, $x \neq 1$ and $y \neq 1$; thus, $x=2$ is the only solution for $x$. Substituting this value into the first equation, we get

$$
\binom{y+1}{3}-\binom{y}{3}=6 \quad ; \quad \frac{(y+1) y(y-1)}{6}-\frac{y(y-1)(y-2)}{6}=6
$$

from which the quadratic equation in $y$, $y^{2}-y-12=0$, follows. The solutions are $y_{1}=4$ and $y_{2}=-3$, but only $y_{1}=4$ lies in the solution domain. Indeed, $x=2$ and $y=4$ satisfy the given system of equations (as a check confirms).