4 different colour balls, each of number four have to be arranged in a circular manner so that adjacent 3 balls are of different colour. I have 4 red balls, 4 green balls, 4 Blue balls and 4 Yellow balls with me. I have to arrange them in a circular manner. The condition is that if we take any 3 adjacent balls, they should have to be of different colors. How many such arrangements are possible?
I have tried fixing the colors Red, Green, and Blue initially. I tried the situation where only two balls of each color are available and wrote all possible combinations. But in the case of 4 balls, I am not able to write all the arrangements or generalize the situation. Can anyone help me with the correct method of doing this problem?
 A: My attempt:
Suppose we've selected a trio at random from a random configuration. What is the probability that at least 2 of them are the same color? Recall we have 4 red, blue, green, and yellow.
If we fix the first ball (color $x$), then it boils down to the choices available for the second two selections to give us what we want, which are as follows:
$$(x,y)\text{ or }(y,x)\text{ or }(x,x).$$
Hence:
$$P(\text{at least 2 are color $x$}) = \frac{4}{16}\bigg(\frac{3}{15}\cdot\frac{12}{14}+\frac{12}{15}\cdot\frac{3}{14}+\frac{3}{15}\cdot\frac{2}{14}\bigg).$$
Since $x$ is an arbitrary color and there are 4 of them, we get:
$$P(\text{"at least 2 are the same color"}) = 4*P(\text{at least 2 are color $x$}).$$
Now take the complementary case:
$$P(\text{all colors are distinct}) = 1 - P("")\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }(\bigstar)$$

We just found the probability that a random trio in a random configuration will satisfy the condition. However, we need all such random trios to satisfy the condition if our configuration is to be admissible. We consider this separately.

For a random trio there are two options, either it satifies the condition or it doesn't. Now, considering there are 16 distinct trios (going by placement of the centers), we have:
$$P(\text{ all trios satisfy the requirement}) = \bigg(\frac{1}{2}\bigg)^{16}\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }(\bigstar\bigstar)$$

As Henry notes, there are $\frac{16!}{(4!)^4}$ configurations total (we can’t distinguish between different balls of the same color). So the number desired is the fraction of this provided by our above probabilities:
$$(\bigstar)*(\bigstar\bigstar)*\frac{16!}{(4!)^4}$$
Simplifying this, I get roughly $604$ such arrangements.
