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

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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 \( 4 * 2 * 2 = 16 \). – Siddhartha Srivastava · 2 years, 3 months ago

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– Calvin Lin Staff · 2 years, 3 months ago

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

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. – Brian Charlesworth · 2 years, 3 months ago

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– Chung Kevin · 2 years, 3 months ago

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

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

Python 2.7:Log in to reply

– Brian Charlesworth · 2 years, 3 months ago

Nice program, Brock. I can't argue with its output; the answer is indeed \(16.\) :)Log in to reply

– Brock Brown · 2 years, 3 months ago

Thanks! :DLog in to reply