To determine number of arrangements of 4 letters in a word so that the transitions remains conserved A 10 letter word is composed of $A,\ B,\ C,\ D$. The problem is to find the number of arrangements of these alphabets which could lead to fixed number of transitions between each pair of alphabets.
Example,  consider the following arrangement of $A,\ B,\ C,\ D$ given as 
$ABACDABCDD$ has $4\ A$, $2\ B$, $2\ C$, $3\ D$ has $3$ transitions between $A$ and $B$, $1$ transition between $A$ and $C$, $1$ transition between $A$ and $D$, $2$ transition between $C$ and $D$ and $1$ transition between $B$ and $C$.
Here the problem is to find the number of ways it can be arranged so that the transition between alphabets remains conserved.
 A: It looks like the answer is 64.  I feel a bit cheap posting this, since my solution doesn't scale and is pretty unilluminating, but sometimes generating functions are easier than thinking.
We'll write down a generating function involving $4$ variables $y_a, y_b, y_c, y_d$ to keep track of individual letters, and $6$ variables $x_{ab},x_{ac},\ldots,x_{cd}$ to track transitions.  The answer will be the coefficient of
$$y_a{}^3 y_b{}^2 y_c{}^2 y_d{}^3 x_{a,b}{}^3 x_{a,c} x_{a,d} x_{b,c} x_{c,d}{}^2$$
in a horrible, horrible polynomial.  (Note there are only $3$ A's, not $4$.)
If we only wanted to keep track of the transitions, we could look at the $9$th power of the matrix
$$A=\left(
\begin{array}{cccc}
 1 & x_{ab} & x_{ac} & x_{ad} \\
 x_{ab} & 1 & x_{bc} & x_{bd} \\
 x_{ac} & x_{bc} & 1 & x_{cd} \\
 x_{ad} & x_{bd} & x_{cd} & 1 \\
\end{array}
\right)$$
whose $(i,j)$ entry is $x_{\min(i,j),\max(i,j)}$, and we've set $x_{ii}=1$ since we don't need to keep track of repetitions of specific letters.  However, we also need to keep track of the counts of individual letters, and since we've symmetrized the transition matrix a reasonable approach is to multiply $A$ on the right by the matrix
$$Y=\left(
\begin{array}{cccc}
 y_a & 0 & 0 & 0 \\
 0 & y_b & 0 & 0 \\
 0 & 0 & y_c & 0 \\
 0 & 0 & 0 & y_d \\
\end{array}
\right)
$$
so that the $y$'s keep track of the letter we're transitioning to.  Note
$$AY=\left(
\begin{array}{cccc}
 y_a & x_{a,b} y_b & x_{a,c} y_c & x_{a,d} y_d \\
 x_{a,b} y_a & y_b & x_{b,c} y_c & x_{b,d} y_d \\
 x_{a,c} y_a & x_{b,c} y_b & y_c & x_{c,d} y_d \\
 x_{a,d} y_a & x_{b,d} y_b & x_{c,d} y_c & y_d \\
\end{array}
\right).$$
Any letter can in principle come first, so we start with the row vector $\mathbf{y}=(y_a,y_b,y_c,y_d)$.  The $j$-th entry in the $4$-long vector $\mathbf{y}(AY)^9$ counts all $10$-long strings ending with letter $j$, each weighted by a product of $y$'s for the letters in the string and a product of $x$'s for the transitions which occur.  The horrible, horrible polynomial is the sum of the terms in the vector; it has $34,702$ terms, but that's tractable with a computer algebra system.  Mine takes about four seconds to tell me that the coefficient of the monomial in question is 64:

