How many ways can you list them if you must alternate students and administrators?

Question

You need to arrange a list of participants - four administrators and four students - who will sit behind a table in the order listed. In how many ways can you list them if you must alternate students and administrators?

This question is from Discrete Mathematics for Computer Scientists by Stien, Drysdale and Bogart.

Here is what I am thinking they be arrange two ways

1) $SASASASA$

or

2) $ASASASAS$

In both $1$ & $2$, you can arrange by $(4*3*2*1)$ students and $(4*3*2*1)$ admins.

My solution would be: $$(4! + 4!) * 2 = 4*4!.$$

The text does not provide a solution, so I hope someone here can confirm.

Answer 1

Your answer is incorrect for several reasons. First, you don't add the totals: since you must order both the students and the administrators, you should multiply the totals; you multiply when you are counting how many ways you can do both $A$ and $B$; you add when you are counting how many ways you can do either $A$ or $B$. See the opening remarks in this previous answer. And you are not taking into account that you have both possibilities: either begin with an administrator, or begin with a student (which would require you to add the results you got for each case).

Here's two different ways of counting them, both leading to the same final answer, both giving the same answer Ross gave, but with details explained.

1. Along the lines you describe, but using the Sum and Product rules correctly: You can either begin with an administrator or with a student. No matter who begins, you order the administrators, which as you correctly point out can be done in $4!$ ways, and you order the students, which again as you correctly point out can be done in $4!$ ways. Since you must to both, in each case (whether an administrator goes first or a student goes first) you have $4!\times 4!$ ways of ordering them (you must order the administrators and order the students, so the Product Rule applies). This describes the full ordering, since once you order the students, order the administrators, and decide who goes first, you've got the entire order.

   Since you can either begin with a student or with an administrator, you want to add the total number of ways to do it with a student first (which is $4!\times 4!$), with the number of ways to do it with an administrator first (also $4!\times 4!$). So the total number is $(4!\times 4!) + (4!\times 4!) = 2\times 4!\times 4!$.

2. You can pick anyone to go first; that's $8$ possible ways. Once you decide who goes first, you've got $4$ possibilities for who goes second (must be someone from the other group). Then you've got $3$ for go who goes third (only three people left in the group to which the first person belongs), and $3$ for who goes fourth (any of the three remaining people in the group to which the person in the second position belongs); then $2$ for who goes fifth, $2$ for who goes sixth, and the 7th and 8th place are now forced. So that gives $$8\times 4\times 3\times 3\times 2\times 2 = 2\times(4\times 3\times 2)\times(4\times 3\times 2) = 2\times4!\times 4!$$ ways of ordering them, same as before.

Answer 2

For each arrangement of students you have $4!$ arrangements of admins, so you have $2$(whether you start with $A$ or $S$) $*4!$ (arrangements of students) $*4!$ (arrangements of admins) $= 1152 $