I'm trying to put some order in my head for this combinatoric problem that I have since I'm finding lot's of ways to get to the awnser but I'm not entirely sure why it is so.
The problem is: How many linear arrangement can you make with the letters A,B,C,D,E and F if
A must be placed before B
I found 3 ways to solve for the problem and I'm wondering which is the most "generalised" or "rigorous" way to solve it
Solution 1
If I place $A$ in the first position, then I can place $B$ in $5$ ways such that $A$ is before $B,$ and for each of the five ways I can place $A$ and $B$ for that case, I can place the rest of the letters in $4!$ ways
Then I do the same reasonning for the rest of the cases (i.e moving $A$ one place to the right implies $4$ ways that $B$ can be placed) and then I end up with
$(4!)5+(4!)4+(4!)3+(4!)2+(4!) = (4!)(5+4+3+2+1) = (4!)(15)$ ways
Solution 2 I don't know if my reasonning stands or if it just happens to work:
I can place $A$ and $B$ in such a way that $A$ is before $B$ in ${6 \choose 2}$
Is that right? I mean can I assume that $A$ will be before $B$?
For each of the ${6 \choose 2}$ ways to place $A$ and $B$, there is $4!$ ways to place the rest of the letters, therefore by the product principle there is
${6 \choose 2}(4!) = 15(4!)$ ways to place the letters
Solution 3:
I can arrange the letters in $6!$ different ways,
for each of these permutations, $A$ is either before or after $B$.
And there's as many ways for $A$ to be before $B$ than for $B$ to be before $A$.
In other words, in half of the permutations (I don't really know how to show this symmetry mathematically and if its the same if for example A would precede B and B would have to precede C), $A$ is before $B$, therefore there is
$\frac{6!}{2}$ ways to place them (which is the same as the other)
So to resume my question would be which one is more "mathematically rigorous" and why are they true, and are my arguments good?
Thank you!!