For All Unique Combinations of 60 A's and 20 B's Number of Combinations that have BB Here is my question. I have 60 A's and 20 B's and need to find out the number of unique combinations of those where B shows up consecutively at least once.
For example (6 A's and 2 B's):
AAAAAABB = 1
AAAAABAB = 0
AAAABBAA = 1
 A: Let's count the ways that $B$ doesn't occur  $2$ or more times in a row.
Line up the $A$'s. They determine $61$ "gaps" (I am counting in the endgaps).
We can choose  $20$ of these gaps to put a $B$ into in $\dbinom{61}{20}$ ways.
Subtract this from the total number of words, which is $\dbinom{80}{20}$.
Added: We want to count the number of good words, where a word is good if it has (somewhere or other) two or more consecutive $B$'s. It is much easier to count the total number of words, which is $\binom{80}{20}$, and subtract the number of bad words, words that nowhere have $2$ or more $B$'s in a row. So we concentrate on counting the bad words. For the sake of illustration, I will assume that there are $12$ $A$'s and $5$ $B$'s. This is because I will be drawing a sort of picture, and don't want to type $60$ $A$'s.
How do we make a bad word, that is, a word of length $17$, with $12$ $A$'s and $5$ $B$'s, and imagine lining up the $12$ $A$'s like this.
$$ A \quad A \quad A \quad A \quad A \quad A \quad A \quad A \quad A \quad A \quad A \quad A 
$$
Where can the $5$ $B$'s go? No two $B$'s can be next to each other, so any $B$ must be placed either in a gap between $2$ consecutive $A$'s or at the left end or at the right end. There are $11$ gaps between consecutive $A$'s, and two end places, which I called endgaps. So there are $11+2=13$ places where the $B$'s can go. To make a bad word, we must choose $5$ of these $13$ places to put a $B$ into. This choosing can be done in $\binom{13}{5}$ ways. 
