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Count finite outcomes with disjoint-case sums, successive-choice products and equal-fiber division or bijections. Check that cases are disjoint and multiplicities are constant. Multiplication of choice counts is not a probability-independence assumption.
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Across the full lesson, the source defines counting through bijections, distinguishes overcounting from undercounting, derives a sequence-count formula and tests a proposed count.
From 36 to 91 seconds, the repeated-letter example removes exchanges within identical groups: 7!/(3!3!)=140. This is finite arrangement counting, not a probability calculation.
A reviewed path from functions and finite counting to arrangements, selections, divisibility and the Euclidean algorithm. Every new concept is paired with verified video evidence, while the prerequisite arrows describe this map’s editorial learning order.
For n distinct objects, successive choices are n, , …, 1, giving n!. The counting rule needs the stated number of choices after each preceding choice; this is not a probability-independence assumption.
Conditions: Objects are distinct.; Order matters.