Number of permutations such that $a-b+c-d+e-f+g-h=0$ All possible permutations $\left\{a,b,c,d,e,f,g,h\right\}$
 of the set $A=\left\{1,2,3,4,5,6,7,8\right\}$ are formed.  How many of those permutations satisfy

$$a-b+c-d+e-f+g-h=0$$

My Try:

we have for example

$$(2-1)+(4-3)+(5-6)+(8-7)=0$$ and each of the number in brackets if we treat them as four letters, they can be arranged in $4!=24$ ways.Now in all these possible permutations if we multiply with negative sign we get a different permutation.

So total is $48$.

similarly for $$(2-3)+(4-1)+(5-6)+(8-7)=0$$ we get $48$ permutations.

but i feel this is an informal approach. Any clue for better approach?
The first thing to notice is that the sum of all the numbers from $1$ through $8$ is $36$ so the numbers that are added must sum to $18$ as must the numbers that are subtracted.  We need to find the number of ways to divide the numbers into two sets of four such that each set adds to $18$.  The $8$ has to go in one set, so we look for ways to have three numbers sum to $10$.  They are $721,631,541,532$ so there are $4$ partitions of the set.  For each partition we have two ways to choose which set is added, $4!$ ways to choose the order of the added set, and $4!$ ways to choose the order of the subtracted set.  This gives a total of $4\cdot 2 \cdot 4! \cdot 4!=4608$
The total of the numbers is $36$.  Divide by $2  = 18$.

So you may have split the numbers in the following groups $= ({1,3,6,8}$ and ${2,4,5,7})$ or (${1,4,6,7}$ and ${2,3,5,8}$) or (${1,4,5,8}$  and ${2,3,6,7})$. 

Each group  could be permuted in $4!$ ways such that $a-b+c-d+e-f+g-h=0$ is true and each could pair could be swapped for add ad subtract and thus there are three pairs and hence 8 to give a total of $ 8\times24\times24 = 4608$