Red, green and blue balls
You have 9 balls - 3 red, 3 green and 3 blue - in a random order.
How many ways can you rearrange them so that no ball is replaced with a ball of the same color?
Assume that similarly colored balls are indistinguishable.
The answer is 56.
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1 solution
Maybe I'm misunderstanding this, but didn't you state in the problem that no ball can be replaced by a ball of a different color? So, the first arrangement defines the three spots for each color that the balls in the rearranged order can lie in, and balls from other colors can't go there.
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Sorry, I goofed. The original wording had a mistake. I've corrected it.
"way" needs to be defined better. For example, are two ways the same if I simply reorder which red ball gets mapped over? IE Is R 1 R 2 R 3 B 1 B 2 B 3 G 1 G 2 G 3 → G 1 G 2 G 3 R 1 R 2 R 3 B 1 B 2 B 3 and R 1 R 2 R 3 B 1 B 2 B 3 G 1 G 2 G 3 → G 1 G 2 G 3 R 2 R 1 R 3 B 1 B 2 B 3 the same way?
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Good point... Hows my update?
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Much better.
The "18 ways that blue can replace two of one color and one of another. And for each of those ways there are 3 ways to place the remaining colors" could be explained in a bit more detail.
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@Calvin Lin – True. Lemme think of a better way to explain....
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@Geoff Pilling – FYI The generalization of the derangements formula to allowing for indistinguishable objects is pretty disgusting. It isn't really illuminating in this case because many of the terms are 0, and the only non-zero terms are these 2 cases that you found.
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@Calvin Lin – Interesting... Lemme see if I can come up with a non-trivial derangement problem...
Thought about suggesting a follow-up where we add 3 yellow balls into the mix. But no .... that becomes an unholy mess. :O
There are 2 ways that one color can be completely replaced by another color.
After that there are 18 ways that blue can replace two of one color and one of another. And for each of those ways there are 3 ways to place the remaining colors.
2 + 3 * 18 = 56