4. How many orderings $\left(a_{1}, \ldots, a_{8}\right)$ of $(1,2, \ldots, 8)$ exist such that $a_{1}-a_{2}+a_{3}-a_{4}+a_{5}-a_{6}+a_{7}-a_{8}=0$ ?
Answer: 4608 We can divide the numbers up based on whether they have a + or - before them. Both the numbers following +'s and -'s must add up to 18. Without loss of generality, we can assume the +'s contain the number 1 (and add a factor of 2 at the end to account for this). The possible 4 -element sets containing a 1 which add to 18 are $\{1,2,7,8\},\{1,3,6,8\},\{1,4,5,8\},\{1,4,6,7\}$. Additionally, there are 4 ! ways to order the numbers following a + and 4 ! ways to order the numbers following $\mathrm{a}-$. Thus the total number of possibilities is $4 \times 2 \times 4!\times 4!=4608$.