Why use C(n,r) instead of P(n,r) when considering how many strings can be formed in which a specific letter appears before another specific letter? I am dealing with a problem in which I must determine how many strings can be formed by ordering the letters ABCDE subject to the conditions given. 
The condition that I am given is that A appears before D, examples of which include BCAED, BCADE.
In tackling this problem, my first instinct was to utilize the permutation formula, specifically P(5,2) because it seems that we are picking 2 out of 5 elements and ORDERING them. Then I proceeded to multiply P(5,2) by 3! i.e. P(5,2) * 3! my reasoning being that after choosing the positions in which the letters A and D appear in the string, there are still 3! ways to order the remaining elements B,C,E in the string. 
The answer however is given as C(5,2)*3!, which is exactly what is confusing me. If we are told that A must appear before D doesn't this imply that order matters and that we are dealing with a permutation? I understand that the order of B,C,E is irrelevant, but don't we care about ensuring that A always comes before D?
Also, when speaking of combinations, aren't we talking about groups in which order does not matter? It just seems to me that if you use C(5,2) you could be implying that you can have A before D but also D before A in positions 1-5 of your string.
Thanks in advance!
 A: It is precisely because we know that $A$ comes before $D$ that we want $\binom52$. The point is that once we know which two positions in the string are occupied by $A$ and $D$, we automatically know which one contains $A$ and which one contains $D$: the earlier one must contain $A$. There are $\binom52$ ways to choose two positions out of the five in the string, and each one of those ways completely pins down the location of $A$ and the location of $D$. 
As you say, the remaining three letters can fill the remaining three slots in any of their $3!$ possible orders, and their order amongst themselves is independent of the choice of the locations for $A$ and $D$, so by the multiplication rule there are altogether $\binom52\cdot3!$ words with $A$ before $D$.
A: To solve the problem using permutations:  Note that there are $5!$ ways to arrange the letters if you ignore the condition.  Transposing $A,D$ shows that there are exactly as many permutations with $A$ first as there are with $D$ first, so the answer is $\frac {5!}2$.
