Ch.17 Permutation and Combination 1 Direct multiplication β Example: If three people A, B and C go to a 3 stories hotel, how many arrangements are there such that A and B would not be on the same floor? ππ’ππππ ππ π€ππ¦π =3×2×3=18 For A is on 1/F, B must be on 2/F or 3/F, vice versa. 2 Permutation β πππ = π! (π − π )! Example: How many ways can Andy, Bob, Cathy be lined up? ππ’ππππ ππ π€ππ¦π = π33 = 6 Note that There are 3 people The “capacity” of the line is 3 We may see it as: π33 Note that Unlike combination, permutation includes arrangement. For example, “Andyο Bobο Cathy” and “Andyο Cathy ο Bob” are different. 3 Combination β πΆππ = π! (π − π )! π! Example: How many ways can Andy, Bob, Cathy form a team of 2? ππ’ππππ ππ π€ππ¦π = πΆ23 = 3 Note that We may see it as: πΆ23 There are 3 people The “capacity” of the team is 2 Note that Unlike permutation, combination excludes arrangement. For example, “Andyο Bob” and “Bobο Andy” are the same. 4 Common question type β Digits Example: How many 3-digit numbers can 0, 1 and 2 form? ππ’ππππ ππ π€ππ¦π = π33 − π22 = 4 012 021 102 120 201 210 012 021 Note that Digits cannot start with 0 Letters and Words Example: How many different words can B, A, L, L form? ππ’ππππ ππ π€ππ¦π = πΆ24 × 2 = 12 Consider the two “L” first: There are 4 spaces to place 2 “L” Lining up Example: If Andy, Bob, Cathy and Daniel line up, find the number of ways such that Daniel is behind Cathy ππ’ππππ ππ π€ππ¦π = π44 ÷ 2 = 12 Note that Daniel is either behind or in front of Cathy Example: There are 2 boys and 3 girls lining up. Find the number of ways such that the boys are standing next to each other. ππ’ππππ ππ π€ππ¦π = 4 × π22 × π33 = 48 π22 π33 _B _B _G _G _G The “box” can be in 4 different position of the 5 spaces Separation Group with less individuals Group with more individuals Example: There are 4 boys and 3 girls lining up. Find the number of ways such that no boys are next to each other. ππ’ππππ ππ π€ππ¦π = π44 × π33 = 144 5 Complementary Cases β π΅πππππ ππ ππππ πππ πππππ π¨ πππ ππ ππππππ = π¨ππ πππππππ − π΅πππππ ππ ππππ πππ πππππ π¨ ππ ππππππ Example: Find the number of ways such that there are at least 1 head in 5 tosses of a fair coin. ππ’ππππ ππ π€ππ¦π = π΄ππ ππ’π‘ππππ − ππ’ππππ ππ π€ππ¦π ππ πππ‘π‘πππ πππ π‘πππ = 25 − 1 = 31