Solution - St John Brebeuf

1MathematicsMath 12 Foundations.DepartmentUnit 4: Counting Methods.

4.1 Counting Principles:2

Show all your working on this page3

4Example 1:Solution:

Number of Uniforms = Number of sweaters . Number of shortsU = 3 . 3U = 95

6Example 2:Solution:

Example 3:8

Solution:9

Summary:11

Homework: Page 235 # (1-17)13

144.2- Introducing Permutations & Factorial Notation:Example 1:

Solution:15

17Example 2:Solution:Use the factorial! button on your calculator

18Use the factorial! button on your calculatorDon’t forget the brackets

19Note:(n+3)! =(n+3) (n+2) (n+1)n(n-1)(n-2)(n-3)(n-4)………….…3.2.1(n+2)! = (n+2) (n+1)n(n-1)(n-2)(n-3)(n-4)………………… …3.2.1(n+1)! = (n+1)n(n-1)(n-2)(n-3)(n-4)…………………………….. 3.2.1n! = n(n-1)(n-2)(n-3)(n-4)……………………………………..3.2.1(n-1)! = (n-1)(n-2) (n-3)(n-4)……………………………………….3.2.1(n-2)! = (n-2)(n-3)(n-4)……………………………………………….3.2.1(n-3)! = (n-3)(n-4)………………………………………………………3.2.1

20Example 3:Solution:

22Example 3:Solution:

25Summary:Homework: Page 243 # (1-16)

264.3-Permutations When All Objects are Distinguishable:Investigate-

27n P r =4 P 3 =To perform this calculation on yourcalculator follow these steps: 4 MathPRE (2) nPr 3 = 24

28n P n or n P nExample 1:

29Solution:Non-CalculatorCalculator10 P 6To perform this calculation on yourcalculator follow these steps: 10 Math (2) nPr 6 = 151 200PRE

30Example 2:Solution:

32Example 3:Solution:

34Perform each calculationindividually then add themtogetherExample 4:

Solution:35

The cars can be parked in 144 different ways36

Example 5:38

39Solution:# 0f digits 0….9 = 10;so there are 10 choices for each spot since repeats are allowed10 10 10 10 10 10 10 10 10

40To perform this calculation on yourcalculator follow these steps: 10 Math (2) nPr 9 = 3 628 800PRE

Summary:41

Homework: Page 255 # (1-18)42

4.4-Permutations When Objects are Identical:43

45Example 1:Solution:

46Example 1:Solution:

Remaining letters are: A, A, D, A47

Example 2:49

Solution:50

There are 8 objects to be arranged5 E’s - E E E E E and 3 S’s – S S S51

52Summary:Homework: Page 266 # 1-18

534.5/4.6- Combinations:Using the letters a, b, and c how many 2 letter permutations can be achieved?ab ba ac ca bc cbThere are 6 two letter permutations ie. 3 P 2 = 6However there are only 3 distinctcombinations:ab ac bcRead as n choose r0 ≤ r ≤ n

54Example 1:Solution:To perform this calculation on yourcalculator follow these steps:9 C 3 == 84 9 Math nCr 3 = 84PRE

55Example 2:Solution:

56To perform this calculation on yourcalculator follow these steps: 10 Math nCr 3 = 120PRE

57Alternative NotationExample 3:

58Solution:To perform this calculation on yourcalculator follow these steps: 5 Math nCr4 = 5PRE

60 5 Math nCr2 = 10PRE4 Math nCr2 = 6PREPerform the 2 calculations separatelythen multiply

61Example 4:Solution:

63Summary:Homework: Page 280 # (1-19)

644.7-Solving Counting ProblemsExample 1:Solution:

Example 2:66

Solution:67

69Example 3:Solution:

71Summary:Homework: Page 288 # (1-17)