Five boxes in a row (positions 1–5) are each filled with exactly one ball, either red (R) or blue (B). No two adjacent boxes may both contain blue balls. Balls of the same color are identical. How many different arrangements (strings over {R, B}) are possible?

Introduction / Context:
We are counting binary strings of length 5 with no consecutive B’s. This classic constraint yields a Fibonacci-type recurrence.

Given Data / Assumptions:

• Length n = 5.
• Alphabet {R, B} with “no BB” adjacency.
• R and B are indistinguishable beyond color (identical balls).

Concept / Approach:
Let a(n) be the number of valid strings of length n. Appending R to any valid string of length n−1 is always allowed; appending B is allowed only to strings of length n−1 that end in R. This leads to a(n) = a(n−1) + a(n−2), a Fibonacci recurrence.

Step-by-Step Solution:
Base cases: a(1) = 2 (R, B); a(2) = 3 (RR, RB, BR).Then a(3) = 5; a(4) = 8; a(5) = 13.Therefore, there are 13 valid arrangements for 5 boxes.

Verification / Alternative check:
Direct enumeration confirms: 8 with at most two B’s spaced apart plus 5 with exactly two B’s non-adjacent.

Why Other Options Are Wrong:
8 and 10 undercount; 15 and 22 overcount by allowing adjacent B’s.

Common Pitfalls:
Assuming each position always has 2 independent choices (which ignores the adjacency ban) or forgetting that RB and BR are both valid.

Final Answer:
13