Question 51
To a sequence of $n$ zeros and $n$ ones, we assign the number of maximal contiguous runs of equal digits in it. (For instance, sequence $00111001$ has four such runs: $00$, $111$, $00$, and $1$.) For a given $n$ we sum up all the numbers assigned to all such sequences. Prove that the resulting sum is equal to
$$
(n + 1) \binom{2n}{n}.
$$
$$
(n + 1) \binom{2n}{n}.
$$