You need to look at thre different cases.
1) The pairs all sit infront/behind.
See how many ways the first row can be filled and the second row will automatically be accounted for.
I think that is just 10*8*6*4*2 = 3840 possibilities.
2) 1 pair sit in each row side by side the others are infront/behind
Side by side couples:
there is 4 possible places to sit the pairs.
There are 10 ways to chose the first person of the first pair and then 8 ways to chose the first person of the second pair
So I guess that is 8*10 *4= 320 Ways.
Now there is 3 paired people/places left 6*4*2=48 ways to sit them
320*48= 15360 possibilities
3) 2 pairs sit in each row, the others are infront/behind.
The individual one can be in the 1st 3rd or 5th positions.
There are 5 ways to choose the odd pair so that make it 3*5*2= 30ways to chose.
Now there is 4 couples to sit side by side. and there are 4 paired places to put them.
4!*2^4=384
30*384 =11520 possibilities
3840+15360+11520 = 30720 possibilities.
Since this answer agrees with your book it has a fair chance of being correct.