10. Compute the number of possible words $w=w_{1} w_{2} \ldots w_{100}$ satisfying:
- $w$ has exactly 50 A's and 50 B's (and no other letters).
- For $i=1,2, \ldots, 100$, the number of $A$ 's among $w_{1}, w_{2}, \ldots, w_{i}$ is at most the number of $B$ 's among $w_{1}, w_{2}, \ldots, w_{i}$.
- For all $i=44,45, \ldots, 57$, if $w_{i}$ is an $B$, then $w_{i+1}$ must be an $B$.
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 $\binom{85}{35}-\binom{85}{34}$