high School⏱ 3 min
Permutations
P(n) = n! arrangements
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6 arrangements of {A,B,C}:
ABC
ACB
BAC
BCA
CAB
CBA
3pos 1
×2pos 2
×1pos 3
= 6Position 1: 3 choices. Position 2: 2 (one taken). Etc. Total P(3) = 3! = 6.
P(n) = n! arrangements
6 arrangements of {A,B,C}:
Position 1: 3 choices. Position 2: 2 (one taken). Etc. Total P(3) = 3! = 6.