## Task B-4.1.

Solve the inequality $\binom{n+1}{6}<\binom{n+1}{n-3}$.
## Solution.

Due to the symmetry of binomial coefficients $\binom{n+1}{n-3}$ is equal to $\binom{n+1}{4}$. 1 point

The condition of the problem is $n \geqslant 5$.

The inequality $\binom{n+1}{6}<\binom{n+1}{4}$ transforms into

$$
\frac{(n+1) n(n-1)(n-2)(n-3)(n-4)}{6!}<\frac{(n+1) n(n-1)(n-2)}{4!}
$$

After simplification, we get

$$
\begin{aligned}
& \frac{(n-3)(n-4)}{30}<1 \\
& n^{2}-7 n-18<0
\end{aligned}
$$

The solution to the inequality is $n \in\langle-2,9\rangle$, 1 point but due to the initial conditions, these are only the natural numbers $n \in\{5,6,7,8\}$.