Suppose a sequence {$a_n$} with $2m$ terms, including $m$ terms equal to $1$ and m terms equal to $0$.
Such a sequence is defined as normal if
For arbitrary $k \leq 2m$, the number of terms equal to $0$ is not less than that of terms equal to $1$ in the first $k$ terms {$a_1, a_2, \ldots, a_k$}.
Otherwise, it is abnormal.
Show that the number of abnormal $0-1$ sequences {$a_n$} with $2m$ terms equals that of sequences {$a_n$} of which $(m + 1)$ terms are $0$s and $(m − 1)$ terms are $1$s.
Below is my thought:
This problem is equivalent to prove $C_{2m}^m - \textbf{number of normal sequences} = C_{2m}^{m-1}$. The total cases of $2m$ sequences with $m$ ones and $m$ zeros minus normal cases should be the number of abnormal cases.
So I tried to find normal cases. I am considering a sequence like $\{0,1,0,1,...,0,1\}$. If I exchange the position of a ${1,0}$ pair, this sequence is still normal. But I failed to go any further. Can anyone help me?