## Task 36/79

Solve the equation

$$
\binom{4 k}{k} \cdot k^{2}=\binom{4 k}{k+1} \cdot(k+1)
$$

with $k \in \mathbb{N}, k>0$, and calculate the value of both terms for the found solution(s).
It is

$$
\binom{4 k}{k} \cdot k^{2}=\frac{(4 k)!k^{2}}{k!\cdot(3 k)!} \quad ; \quad\binom{4 k}{k+1} \cdot(k+1)=\frac{(4 k)!(k+1)}{(k+1)!\cdot(3 k-1)!}
$$

Therefore, the given equation can be written in the form

$$
\frac{(4 k)!k^{2}}{k!\cdot(3 k)!}=\frac{(4 k)!(k+1)}{(k+1)!\cdot(3 k-1)!}
$$

Now, it can be further equivalently transformed:

$$
\begin{array}{r}
\frac{k^{2}}{k!\cdot(3 k)!}=\frac{(k+1)}{(k+1)!\cdot(3 k-1)!}=\frac{1}{k!\cdot(3 k-1)!} \\
k^{2}=\frac{(3 k)!}{(3 k-1)!}=3 k
\end{array}
$$

From this, it immediately follows that $k=3$ because $k>0$. Substituting this into the given equation, we get

$$
\binom{12}{9} \cdot 9=\binom{12}{4} \cdot 4 \quad ; \quad \frac{12 \cdot 11 \cdot 10}{1 \cdot 2 \cdot 3} \cdot 9=\frac{12 \cdot 11 \cdot 10 \cdot 9}{1 \cdot 2 \cdot 3 \cdot 4} \cdot 4=1980
$$

### 2.20 Problems and Solutions 1980