I could only do q10 with a brute force approach/ I'd never get this and there's probably a better way I can't think of right now.
So:
1\cdot (n\cdot n), n\in\{2, 3, 4, 5, 6\} \Rightarrow 3\cdot 5=15 different permutations.
Now, let's consider all the square number that are available: 1, 4, 9...