You have a standard 8×8 chessboard (64 squares) and 10 distinguishable coins. You are creating sequential arrangements by placing coins on the board under these rules: 1. In each arrangement, you place 1 to 10 coins on the board, one coin per square, no two coins on the same square. 2. **Across all arrangements combined**, a given coin can only occupy a given square **once ever**. For example, if coin A is placed on square 17 in arrangement #1, coin A can never be on square 17 again in any future arrangement. 3. Each arrangement must place at least 1 coin. What is the maximum number of valid arrangements you can create before it becomes impossible to make a new one? Provide your mathematical reasoning, strategy for maximizing arrangements, and the final answer.
Model the problem
Think of a coin–square pairing as a resource.
There are (10) distinguishable coins and (64) squares, so there are
[ 10\cdot 64 = 640 ]
distinct coin–square pairs ((\text{coin } i, \text{square } s)).
Rule (2) says: each pair can be used at most once total across the whole sequence.
In one arrangement, if you place (k) coins, you consume exactly (k) distinct coin–square pairs. Also within that arrangement, no two coins may share a square (rule 1), but that only constrains simultaneous placements, not reuse by different coins later.
So the total number of arrangements is maximized by making each arrangement use as few new pairs as possible—i.e. use (k=1) coin per arrangement.
That immediately gives an upper bound:
- Each arrangement uses (\ge 1) unused coin–square pair.
- There are only (640) such pairs.
So:
[ \text{#arrangements} \le 640. ]
Achievability (construction)
We can actually reach (640) arrangements by using exactly one coin each time.
For example:
- For coin 1: place it on each of the 64 squares once (64 arrangements).
- Then coin 2: place it on each of the 64 squares once (another 64).
- Continue through coin 10.
This yields (10\times 64 = 640) arrangements.
All rules are satisfied:
- Each arrangement places 1 coin (allowed: 1 to 10).
- No two coins share a square (trivially, since only one coin is placed).
- No coin repeats a square, since each coin visits each square at most once.
After these (640) arrangements, every coin has used all 64 squares, so no coin can be placed anywhere without violating rule (2). Hence no further arrangement is possible.
Final answer
[ \boxed{640} ]
Maximum number of valid sequential arrangements: 640.
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