## Task B-4.1.

Solve the equation

$$
\binom{x+1}{x-2}+2\binom{x-1}{3}=7(x-1)
$$
## Solution.

If we apply the rule $\binom{n}{k}=\binom{n}{n-k}$ to the expression $\binom{x+1}{x-2}=\binom{x+1}{x+1-x+2}=\binom{x+1}{3}$, the given equation transforms into the form $\binom{x+1}{3}+2\binom{x-1}{3}=7(x-1)$.

The condition for the problem to be well-defined is $x+1 \geqslant 3 \Longrightarrow x \geqslant 2$ and $x-1 \geqslant 3 \Longrightarrow x \geqslant 4$.

Therefore, for the solution to be valid, $x \geqslant 4$ must hold.

By applying the rule for calculating binomial coefficients, we get

$$
\begin{gathered}
\frac{(x+1) x(x-1)}{6}+2 \frac{(x-1)(x-2)(x-3)}{6}=7(x-1) / \cdot \frac{6}{x-1} \\
x^{2}+x+2\left(x^{2}-3 x-2 x+6\right)=42 \\
x^{2}-3 x-10=0 \quad \Longrightarrow \quad x_{1}=5, x_{2}=-2
\end{gathered}
$$

Since $x \geqslant 4$, the solution to the given equation is $x=5$.