# Placing square on a grid - Part 3 On a 3 by 3 grid, the red, yellow, blue and green 2 by 2 squares are placed in some order. They have to be placed in their location (IE not moved somewhere else). There are $4 ! = 24$ different ways to place these squares.

Of these 24 different ways, how many distinct results would we get?

For example, placing 1432 and 4132 will both yield  Note by Chung Kevin
5 years, 5 months ago

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$16$.

I'll be referring to the 2 X 2 squares as numbers, and each of the 9 squares as blocks. First, look at the four corner blocks. They can only be covered by the number on it. So we can disregard those. We are left with a central cross. The center of this cross can be covered in $4$ ways, since whichever number we choose first covers it. Now, we are left with the four other blocks. Of these, $2$ will be covered by the first number we chose, so they don't matter. We are left with two blocks. Each of these blocks can be covered by two numbers and each choice is possible, so we have $4 * 2 * 2 = 16$.

- 5 years, 5 months ago

I'm getting a total of $20,$ although your reasoning seems sound. As I understood the question the first number forms the bottom "layer", the second number the second layer and so on, so that the last number/color is on top. For the sequences that end in $23, 32, 14, 41$ it doesn't matter what the order of the first two layers is, so this produces $4$ duplicates. For the other $8$ possible pairs of top layers it will matter how the first two layers are chosen, so there are no duplicates here. Thus I get $24 - 4 = 20$ sequences that produce distinct patterns. Where is the flaw in my logic?

Edit: O.k., I see now. $1324$ gives the same result as $1234$, $4321$ gives the same result as $4231$, $3142$ gives the same result as $3412$ and $2143$ gives the same result as $2413.$ This gives us $4$ more duplicates, leaving us with $16$ distinct outcomes, as Siddhartha and Brock have found.

- 5 years, 5 months ago

Wow, that's a nice analysis! It's certainly simpler than I thought it would be.

Staff - 5 years, 5 months ago

Nice approach. I did it in a similar manner too :)

- 5 years, 5 months ago

I'm finding $\boxed{16}$, gentlemen.

Python 2.7:

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 from itertools import permutations found = set() class Grid(): def __init__(self): self.grid = [['w']*3,['w']*3,['w']*3] def __str__(self): string = '' for row in self.grid: for color in row: string = string + color string = string + '\n' return string def paste(self,origin,color): for x in xrange(2): for y in xrange(2): self.grid[origin+x][origin+y] = color position = {'r':(0,0),'g':(1,1),'b':(1,0),'y':(0,1)} for colors in permutations('rgby'): grid = Grid() for color in colors: grid.paste(position[color],color) found.add(str(grid)) print "Answer:", len(found) 

- 5 years, 5 months ago

Nice program, Brock. I can't argue with its output; the answer is indeed $16.$ :)

- 5 years, 5 months ago

Thanks! :D

- 5 years, 5 months ago