Permutations: (When all objects are distinct)

This section builds up to the formal idea of a permutation using small, concrete seating examples before the general theorem is proved in the next sub-section.

The first example asks: in how many ways can 3 persons be seated on 3 numbered chairs in a row? Labelling the chairs 1, 2, 3 and the persons A, B, C: the 1st chair can be filled in 3 ways (any of the 3 persons); having filled it, the 2nd chair can be filled by either of the 2 remaining people; the 3rd chair is then filled in exactly 1 way, by whoever is left. By the Multiplication Principle, the total is $3\times2\times1=3!=6$ seatings — a count that can be directly verified by listing all six: ABC, BAC, CAB, ACB, BCA, CBA. Extending this reasoning from 3 persons to $n$ persons seated in $n$ chairs in a row gives $n!$ total arrangements. A closing Note stresses the two conditions that make this reasoning valid: all persons must be genuinely distinct from each other, and the chairs must have their own fixed ordinal identity (1st, 2nd, 3rd, ...) — both of these assumptions will later be relaxed (identical objects; circular rather than row seating) in later sub-sections.

A second example changes the numbers: 4 persons and only 2 chairs in a row. The 1st chair can be filled in 4 ways; having filled it, the 2nd chair can be filled in 3 remaining ways; total $=4\times3=12$. This is again confirmed by an explicit listing, grouped by who occupies the first chair: AB, AC, AD (starting with A); BA, BC, BD (starting with B); CA, CB, CD (starting with C); DA, DB, DC (starting with D) — 12 pairs in total.

Extending the argument once more: with 7 distinct objects available and 3 boxes arranged in a row, one object per box, the number of ways is $7\times(7-1)\times(7-2)=7\times6\times5$ — the same shrinking-count pattern seen in both smaller examples, now with 3 factors instead of 2 (since there are 3 boxes/positions to fill). …