, 2017

Solution:
Call the last property in the problem statement $P(i, j)$ where in the statement $i = 44$, $j = 57$. We show that the number of words satisfying the first two conditions and $P(m, m + k)$ is the same independent of $m$ (assuming $k$ is fixed). It suffices to show that the number of words satisfying $P(m - 1, m + k - 1)$ is the same as the number of words satisfying $P(m, m + k)$. Construct a bijection as follows: for a word satisfying $P(m - 1, m + k - 1)$, if it satisfies $P(m, m + k)$, leave it as is. Otherwise, the character in position $m + k$ must be $B$ and the character in position $m + k + 1$ must be $A$. In this case, move these two characters to positions $m - 1, m$ (shifting all other characters back). It is not difficult to verify that this is indeed a bijection.

Thus, the problem is now equivalent to computing the number of words satisfying the first two conditions and $P(1, 14)$. However, this condition simply means that the first $15$ characters must be $B$. Now we are essentially counting the number of paths from $(15, 0)$ to $(50, 50)$ that don't go above $y = x$. There is a bijection between paths from $(15, 0)$ to $(50, 50)$ that do cross $y = x$ and paths from $(15, 0)$ to $(49, 51)$ (using the standard reflection argument). Thus the answer is $\binom{85}{35} - \binom{85}{34}$.