Combinatorics

Permutations

P(n) = n! arrangements

  • High school
  • about 3 min
  • 6 of 11 in the family
Not started

6 arrangements of {A,B,C}:

ABC
ACB
BAC
BCA
CAB
CBA
3pos 1
×
2pos 2
×
1pos 3
= 6
P(3)=3!=3×2×1=6P(3) = 3! = 3\times2\times1 = 6

Position 1: 3 choices. Position 2: 2 (one taken). Etc. Total P(3) = 3! = 6.

Where it is taught