Task B-4.1. (20 points) Solve the system of equations (in the set of natural numbers)

$$
\left\{\begin{array}{l}
\binom{x}{y}=\binom{x}{y+2} \\
\binom{x}{2}=153
\end{array}\right.
$$
Solution. From $\binom{x}{2}=153$ we get $\frac{x(x-1)}{2}=153$ i.e. $x^{2}-x-306=0$. The solutions to this equation are $x_{1}=18, x_{2}=-17$. Since we are only looking for solutions in the set of natural numbers, the only possible solution is $x_{1}=18$.

(10 points)

First method. Now from the first equation we have

$$
\binom{18}{y}=\binom{18}{y+2} \quad \Longrightarrow \quad \frac{18!}{(18-y)!y!}=\frac{18!}{(16-y)!(y+2)!}
$$

which means

$$
\begin{aligned}
(18-y)(17-y) & =(y+1)(y+2) \\
38 y & =304
\end{aligned}
$$

i.e., $y=8$.

(10 points)

Second method. From the symmetry property of binomial coefficients, we have

$$
y+y+2=x
$$

from which $y=8$.