## Task B-4.1.

Solve the equation

$$
\binom{n}{n-2}+2\binom{n-1}{n-3}=\binom{n+1}{n-1}+\binom{n-2}{n-3}
$$
## Solution.

From the symmetry of binomial coefficients, it follows that

$$
\binom{n}{2}+2\binom{n-1}{2}=\binom{n+1}{2}+\binom{n-2}{1}
$$

The solution can be a natural number \( n \geqslant 3 \).

By expanding the binomial coefficients, we get:

$$
\begin{aligned}
& \frac{n(n-1)}{2}+2 \cdot \frac{(n-1)(n-2)}{2}=\frac{(n+1) n}{2}+(n-2) \quad / \cdot 2 \\
& n^{2}-n+2\left(n^{2}-2 n-n+2\right)=n^{2}+n+2 n-4 \\
& n^{2}-5 n+4=0
\end{aligned}
$$

The possible solutions to the equation are \( n_{1}=4 \) and \( n_{2}=1 \).

Given the condition, the only solution is \( n=4 \).