To solve this, you want to line up 2 red balls (R) and 1 blue ball (B) in a way such that there are 3 blocks, which means each ball forms its own block and no consecutive balls have the same color. This ensures no two same-color balls are adjacent.

Key Observation:

To achieve 3 blocks, the balls must alternate colors because any two consecutive balls of the same color will merge into one block. Hence, the only valid arrangement is:

$R \, B \, R \quad \text{(or equivalently)} \quad B \, R \, B$

However, since there are only 2 red balls and 1 blue ball, the arrangement $B \, R \, B$ is not possible because it requires 2 blue balls.

Therefore, the only possible arrangement is:

$R \, B \, R$

Step-by-step Explanation:

1. Fix the arrangement $R \, B \, R$:

   • The first position must be red.
   • The second position must be blue.
   • The third position must be red.

2. Count the arrangements:

   • Since the red balls are indistinguishable, there is only 1 way to arrange the balls to satisfy the condition.

Final Answer:

$\text{There is exactly **1 way** to line up the balls to make 3 blocks.}$

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Let me know if you'd like further details! Here are five related questions for practice:

1. What if you had 2 red balls and 2 blue balls? How many ways could you arrange them to make 4 blocks?
2. How many ways can you line up 3 red balls and 2 blue balls to create exactly 3 blocks?
3. If you had 4 balls of different colors, how many arrangements create 4 blocks?
4. What if the balls were distinguishable? How would that affect the number of ways to create 3 blocks?
5. What is the total number of ways to arrange 3 balls (2 red, 1 blue) without any block constraints?

Tip: When solving block-related problems, consider how "adjacent" elements merge to simplify counting!