Exercise 1: Awarding Gold, Silver, and Bronze Among 8 Racers — Possible Solution
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IDENTIFYING WHICH FORMULA APPLIES
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This is a permutation problem, not a combination problem - per this
chapter's own guiding question, "does the arrangement or order of the
selected items actually matter for this problem?" Here it clearly
does: winning gold versus winning bronze are genuinely different
outcomes, even for the exact same three racers. Choosing which 3
racers place, AND which specific medal each of them gets, means order
matters.

SETTING UP THE CALCULATION
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This is P(n, r) with n = 8 (total racers) and r = 3 (medal positions
being filled):

  P(8, 3) = 8! / (8 - 3)!
          = 8! / 5!

COMPUTING THE RESULT
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Rather than computing the full factorials, the 5! in the denominator
cancels with the bottom of 8!, leaving only the top 3 factors:

  8! / 5! = 8 x 7 x 6 x 5! / 5! = 8 x 7 x 6 = 336

So there are 336 different ways to award gold, silver, and bronze
among 8 racers.

WHY THIS INTUITIVELY MAKES SENSE AS A CHAIN OF CHOICES
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This also follows directly from this chapter's own multiplication
principle: there are 8 choices for who gets gold, then 7 remaining
racers to choose from for silver, then 6 remaining for bronze -
8 x 7 x 6 = 336, exactly matching the formula's own result. The
factorial formula and this direct "chain of diminishing choices"
reasoning always agree, since that's precisely what n!/(n-r)! is
computing.

WHY THIS WORKS AS AN ANSWER
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It explicitly justifies why this is a permutation rather than a
combination using this chapter's own guiding question, sets up and
computes the formula correctly, and cross-checks the result using an
independent multiplication-principle argument that arrives at the
same number.