Problem 15 · 2017 AMC 8
Medium
Counting & Probability
careful-countinggrid
In the arrangement of letters and numerals below, by how many different paths can one spell AMC8? Beginning at the A in the middle, a path allows only moves from one letter to an adjacent (above, below, left, or right, but not diagonal) letter. One example of such a path is traced in the picture.
8C88CMC8CMAM8CMC8C8
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Answer: D — 24 paths.
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Hint 1 of 2
The grid is fully symmetric, so don't trace 24 separate paths. Just ask: from a letter, how many choices for the next letter? Each step's choices are the same no matter which path you're on.
Still stuck? Show hint 2 →
Hint 2 of 2
That's the multiplication principle: when each stage offers an independent number of choices, multiply them. Count the fan-out A→M, then M→C, then C→8.
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Approach: multiplication principle on the fan-out
- From the center A: 4 adjacent M's (up, down, left, right) — 4 choices.
- From any M: 3 adjacent C's (the fourth neighbor is the A you came from, which doesn't extend the spelling) — 3 choices.
- From any C: 2 adjacent 8's — 2 choices.
- Because every stage's choice count is independent of earlier picks, multiply: 4 × 3 × 2 = 24.
- Why this transfers: spell-a-word and lattice-path counts almost always collapse to multiplying the branch count at each step — far faster than drawing every route.
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