I'm discussing with a friend about a solution to a probability question: What is the probability of getting the same number on two symmetric dice and a different number on three other dice (that number is the same on these three dice)?
My answer is ${6\choose{2}} \cdot {5\choose{2}} \cdot \left(\frac{1}{6}\right)^5$. I choose two dice on which the number is the same (so the other three are also determined) and two different numbers out of six, and number of all possibilities is $6^5$. Can you correct me if I'm wrong?
Your answer is not quite right.
Suppose we have five different color dice available. Choose three of those five colors for the triple. This also determines the double. There are six possible values for the number that appears on the triple, which leaves five possible values for the number that appears on the double. Hence, the number of possible outcomes is $$\binom{5}{3} \cdot 6 \cdot 5$$ so the desired probability is $$\frac{\binom{5}{3} \cdot 6 \cdot 5}{6^5}$$ The reason your answer is incorrect is that it matters whether you have a 4 on the triple and a 3 on the double or a 3 on the triple and a 4 on the double. When you simply choose two values for the triple and the double, you fail to distinguish between those cases. Hence, your answer is half what it should be.