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第 20 週 · 排列組合與概率 · 第 2 節

排列:把東西排成一排(nPr)

從一堆不同的東西裡,取幾個「排成一排」—— 順序很重要,ABC 和 BCA 算不同。這節用「填位置法」「捆綁法」把排列數 nPr 與階乘 n! 講透,是 DSE 排列組合的核心招式。配滿圖例。

▶ 第一關: 排列 = 有順序地排(填位置法)
order matters: ABC is not BCAABC= one wayBCA= different waypos 1pos 2pos 3
fill positions one by one (5 people, take 2)pos 15pos 24×5 choices1 used, 4 left5 × 4 = 20 ways
3 letters take 2: tree = 3 × 2 = 6ABCABACBABCCACB6 ways
▶ 第二關: 階乘 n!(全排列)
4 people fill 4 seats = 4!4321×××4! = 4 × 3 × 2 × 1 = 24
factorial grows fast1! = 12! = 23! = 64! = 245! = 1206! = 7207! = 5040each step × a bigger number
▶ 第三關: 排列數 nPr 公式
nPr = n! / (n − r)!= n × (n−1) × … (r factors)start at n, count down, stop after rnn-1...r factorsr positions = r numbers multiplied
5P2 worked out54×=205!/3! cancels 3·2·1, leaves 5 × 4
9P3: count down 3 from 9987××=5043 positions → exactly 3 factors
⚔ BOSS: 排列綜合(相鄰 / 定位限制)
adjacent: bundle the 2 girls firstMMMFF1 unitnow 4 units in a row: 4! = 24girls inside the bundle: 2! = 224 × 2 = 48
fixed position: A must be at the frontA4321fixed1 × 4! = 4 × 3 × 2 × 1 = 24