How do I solve this problem...? Placing 3 coins in a 4x4... I have this math problem:
Three coins are randomly placed in three distinct squares of the 4 by 4 grid
below. What is the probability that no two of them are in the same row or in
the same column or on the same big diagonal (four cells diagonal)?
So, I started to think about it by saying where can we place the first coin? Well, it can go in the corners, Middle 4 squares, and the edges. Then, where can we place the next coin? Well, if it goes in a corner, there is a 1/16 possibility of choosing a corner, then 9 squares are blocked off, and the possibility of choosing another square is 6/15, then the third coin has only one place to go.
I kept using this logic to end up with the following possibilities:
First starts in corner = 1/420
First starts in edge = 6/420
First starts in middle 4 squares: 6/420
Thus, the total probability is 13/420. Apparently that is wrong. Does anyone have a better solution to this?
 A: First off, there can be at most one coin in a corner (each corner prohibits any of the other corners from being used). And there can be at most one coin in the middle 4, for the same reason.
Now, let's count possibilities:


*

*No coin in corners, no coin in middle. Then there are only four pairs of adjacent edge cells to choose from. Only one coin can go in each pair, so we must choose three of the pairs. This can be done in $4$ ways, and two of the three pairs must be opposite. That means $4$ ways to place the three coins in the chosen three pairs. So $16$ ways in total.

*One coin in corner, no coin in middle. There are $4$ choices for the corner. After that, there are only two available pairs of edges, so one coin must go in each such pair, which can be done in $4$ ways. So in total $16$ ways.

*No coin in corners, one in middle. There are $4$ middle cells to place the first coin. Once you've placed it, there are four remaining valid edge cells, but the two remaining coins can't be opposite one another, so there are in total $4$ ways to place the remaining $2$ coins (one either on the left side or the right side, and one either on top or on bottom). In total $16$ ways. 

*One coin in corner, one coin in middle. There are $4$ corners to choose from, and after a coin has been placed in a corner there are $2$ remaining middle cells to place the second coin. Finally, there are only $2$ remaining edge cells available to the final coin (the corner coin taking away two pairs, and the middle coin taking away one cell from each of the remaining two pairs), so all in all we get $16$ possibilities.


All in all, this turns out to be $64$ valid ways of placing the three coins. Out of a total of $\binom{16}{3} = 560$ ways of placing the three coins without restrictions, this means that the probability that randomly placing the coins will produce a valid configuration is $\frac{64}{560} = \frac{4}{35}$.
