TITLE: How many possible words of this type can be formed?
QUESTION [5 upvotes]: We are making $10$ letter words using the letters $A,C,G,T$.
How many possible words are there of the form $A...AC...CG...GT...T$ 
This is where all of the $A's$ go before the all of the $C's$ and all of the $C's$ go before all of the $G's$ and all of the $G's$ go before all of the $T's$?
I know there are $4^{10}$ possible words comprised of the letters $A,C,G,T$ but I have no idea how to account for the overcounting in this bit.

REPLY [8 votes]: This problem is equivalent to finding the number of ways to partition $10$ into four parts, in which the order of the partitioning matters. For example, $10 = 2 + 3 + 3 + 2$ is not the same as $10 = 3 + 2 + 2 + 3$. The first number represents the number of A's, the second number represents the number of C's, and so on.
In order to then count the number of ways, think of $10$ dots and $3$ lines. The $3$ lines will divide the $10$ dots into four parts. So, we have a total of $10 + 3 = 13$ spaces to choose from, and we need to place 3 lines. Therefore, the answer will be ${13 \choose 3} = 286$.
Illustrations of $10$ dots and $3$ lines:
$$\cdot\cdot|\cdot\cdot\cdot|\cdot\cdot\cdot\cdot|\cdot$$
$$\cdot\cdot\cdot|\cdot|\cdot|\cdot\cdot\cdot\cdot\cdot$$
$$\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot|\cdot\cdot||$$
$$\cdot\cdot\cdot|\cdot\cdot\cdot|\cdot\cdot|\cdot\cdot$$

REPLY [7 votes]: This is "stars-&-bars" - you're picking the points in the string where the 3 transitions are. So it's just $${10+3 \choose 3}=\frac{13!}{10!\:3!} = 286$$

Effectively we're extending the ten real positions in the string to include an extra three positions to place the transition markers (the "bars"). 
$$\circ\circ\circ\circ\circ\circ \circ\circ\circ\circ\circ\circ\circ$$
The transition markers are in predetermined order so they do not need identification; we can mark them up after we have selected their positions:
$\newcommand{transit}[2]{\tiny{\frac{#1}{#2}}}$
$$\circ\circ\circ/\circ\circ \circ\circ//\circ\circ\circ$$
$$\circ\circ\circ{\transit AC}\circ\circ \circ\circ{\transit CG}{\transit GT}\circ\circ\circ$$
Then can fill in the real values - the "stars" - based on the separators.
$$AAA{\transit AC}CCCC{\transit CG}{\transit GT}TTT$$
and eliminate the transitions:
$$AAACCCCTTT$$
It's an easy and common blind alley to explore, to look at all combinations and then try to think of a way to eliminate duplicates, but this is obviously a far simpler approach.