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.
 A: Using a smaller example, I expect that rotations of the square are considered identical.  E.g. with $5$ numbers, $\begin{bmatrix}&1&2&\\&&&3\\&&&4\\&&5\end{bmatrix}$ is considered to be the "same" as $\begin{bmatrix}&&&\\&&&1\\
5&&&2\\&4&3\end{bmatrix}$ but not the same as $\begin{bmatrix}&&1&\\&&&2\\&&&3\\&5&4\end{bmatrix}$

Temporarily choose the top left corner to be special, "cut" and "unwrap" the available positions into a line of $20$ available spaces.


*

*Choose which eight of the positions in the line will be occupied by empty space:  $\binom{20}{8}$ options

*From left to right, for each remaining space, choose one of the remaining unused numbers to occupy that space.  $12!$ options.


Paste the line back into position around the square.
Recognize now, however, that every arrangement has been overcounted, each a total of four times.  To make it so that we counted each arrangement once, instead of four times, we divide by four to account for symmetry.  I.e. we "forget" which corner we had made the initial cut at.
The total is then $\binom{20}{8}\cdot 12!\cdot\frac{1}{4}=20\cdot 19\cdot 18\cdots 10\cdot 9\cdot\frac{1}{4} = 5\cdot ~_{19}P_{11}$

The answer you suggested, $\binom{12}{5}\binom{7}{5}\cdot 2\cdot 5!$... this will count the number of ways in which you can take twelve balls labeled 1 through 12, pull five of the balls to place in a bag, pull five of the remaining balls to place in a line, and pull one of the remaining balls to put in your pocket.
This does not count what you hoped for a number of reasons: There might not necessarily be five numbers on a side.  There could be fewer.  The order of the numbers matters on all sides, not just on one of the sides.  There are four sides to consider, you seem to have run out of numbers before reaching the fourth side.
A: As pointed out by JMoravitz, the question is somewhat ambiguous.  However here is a way of justifying the answer you were given.
There are $20$ places available.  Choose one place to take the number $1$, one of the remaining $19$ to take the number $2$ and so on.  The number of possibilities is
$$20\times19\times\cdots\times9=20P(19,11)\ .$$
However, if you rotate the whole arrangement through $0,1,2$ or $3$ right angles, you will get an arrangement which "looks the same".  So the number of "really different" arrangements is a quarter of the above, that is,
$$5P(19,11)\ .$$
A: 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. 
