Arranging numbers around a square In how many ways numbers 1 to 12 can be arranged on a sides of squares (5 places on each sides i.e 20 places total) leaving 8 places empty?
I am getting answer as 12c5(selecting 5 numbers)*7c5(selecting another 5 numbers)*2(remaining 2)*5!(arranging the 5 numbers amongst themselves)
But the answer given is 5*19P11..
Can someone please help me.

I think the reason your answer is wrong is because you assumed that two sides would be filled on the square and a third side would have two numbers. For example, imagine a square with 3 numbers filled on each side with two numbers empty on each side. You did not account for such a possibility in your computation.
The simple explanation behind the answer given is you fix one side of the square to account for rotations and then choose one of the five places on that fixed side for there to be the fixed number. The 19P11 is the number of ways to select and order the remaining 11 numbers in 19 remaining spots. 
I would calculate it actually as 20P12*1/4 which is the number of ways to choose 12 spaces and order 12 numbers in those spaces divided by 4 to account for each rotation being overcounted exactly four times each.
EDIT: The difference between my calculation and the given one is the in which we choose to address rotations being the same. The given answer fixes rotations by automatically choosing a side and number to account for rotations. For the people who are not familiar with fixing rotations in such a manner for non-circle like objects, the safe route would be simply to overcount and then correct for overcounting by multiplying by 1/4, which you can be sure of. 
