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:
different permutations.
Now, let's consider all the square number that are available: 1, 4, 9, 25, 36, 64, 81, 100, 121, 144.
If we break down each square number into factors:
- [+1]
- [+3] - Three different ways to make this.
- [+0] - We've already counted this.
- [+3] - Three different ways for each different combination.
- [+0] - We've already counted this.
- [+6] - Three different ways for each different combination.
- - Not possible
- [+1]
- - Not enough places to make anything.
- [+3] - Three different ways to make this.
- - Not possible
- [+3] - Three different ways for each different combination.
Total:
Total number of permutations:
Since this is higher than all other options, then there must be a few we missed later with higher numbers.
Hence, D