I am faced with organizing a team of 5 speakers: $A,B,C,D,E$.
I must figure out how many permutations I can organize the five speakers, such that $Speaker\,A$ presents before $Speaker\,B$.
A valid order of speakers, which fulfills the above condition, is $A,B,C,D,E$.
Through methods used in other online forums, I stumbled upon the (possible) solution for this answer. From what I can tell, it uses the rule of sum and indirect method calculations to deduce how many permutations are possible ($n$) in these conditions:
$n= 4!+(4!-3!)+(4!-2(3!))+(4!-3(3!))$
$n=4(4!)-6(3!)$
$n=60$
I have tried to look online for potential matches to my technique, such as in falling/rising factorials, although the summations and algebra looks very different from what my solution provides.
Should my solution be correct, what is the mathematical technique being used here and how does it actually work to solve the above problem?
The simplest is to note that there are $5!$ ways to order the speakers. You can pair up each order where $A$ precedes $B$ with a matching one where $B$ precedes $A$, so the number where $A$ precedes $B$ is $\frac 12 \cdot 5!=60$
Another way is that there are ${5 \choose 2}=10$ ways to choose the slots for $A$ and $B$. Then you must put $A$ in the earlier one. There are $3!=6$ ways to put the other people in the $3$ remaining slots.