Let 
be a two letter alphabet. Let 
denote the set of words on this alphabet -- that is to say
finite sequences each entry of which is drawn from the alphabet.
The number of terms in a sequence is its length. Notice that
the number of words of length n is 
since each of the n terms
may be chosen independently. The number of words involving exactly
r copies of B and s copies of W is 
This
quantity is known as the binomial coefficient 
and arises in algebra via
where x and y are unknowns. These quantities 
also form the entries of Pascal's triangle where
r+s is the row number and r indexes how far the entry is from the
left (and s from the right). Note that 
in order to make everything
work smoothly. Putting x=y=1 in (1) yields that
there are 
words of length n, a fact which we have already noted.
The numbers from Pascal's triangle crop up frequently when performing
enumeration arguments.
We will need a fact about these binomial coefficients. Notice that
because then 
and the binomial theorem applies.
This really says that if you add up the entries in any row of
Pascal's triangle (except the top one), alternating the signs
as you go, the answer is 
