## Task 27/72

We are looking for all natural numbers $m$ and $n$ for which the proportion holds:

$$
\binom{n+1}{m+1}:\binom{n+1}{m}:\binom{n+1}{m-1}=5: 5: 3
$$
We first consider the first of the proportions:

$$
\binom{n+1}{m+1}:\binom{n+1}{m}=\frac{(n+1)!m!(n-m+1)!}{(m+1)!(n-m)!(n+1)!}=\frac{n-m+1}{m+1}=1
$$

From this, it follows that $n=2m$. The second proportion yields

$$
\binom{n+1}{m}:\binom{n+1}{m-1}=\frac{(n+1)!(m-1)!(n-m+2)!}{m!(n-m+1)!(n+1)!}=\frac{n-m+2}{m}=\frac{5}{3}
$$

Substituting $n=2m$ into this last relation, we get $\frac{m+2}{m}=\frac{5}{3}$, so $m=3$ and thus $n=6$. As the solution process shows, this is the only solution.