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Counting Principle with Special Conditions

Learn how to apply the Counting Principle with special conditions, such as repetition and without repetition. Understand the concept of permutations and how to calculate the number of possible arrangements.

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Counting Principle with Special Conditions

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  1. Notes #22 The Counting Principle with Special Conditions

  2. Fund. Counting Principle with Repetition Ohio License plates have 3 #’s followed by 3 letters. EX 1: How many different license plates are possible if digits and letters can be repeated? 6 Steps • # of Events: a) Event 1: Numbersb) Event 2: Letters 2) Items per Event a) Event 1: 10 Digits b) Event 2: 26 Letters 3) Indicate the Conditions License Plate Format (3 #’s for Event 1, then 3 letters for Event 2) 4) Set up the“blanks” needed for each Event & fill in the # of items. Event 1: 3 #s Event 2: 3 Letters _____ _____ _____ ______ ______ ______ 5) Multiply the Items: 10*10*10*26*26*26= 17,576,000 different plates 6) Concl Sentence: 17,576,000 different plates can be made if the digits and letters can be repeated. 10 10 26 26 10 26

  3. Fund. Counting Principal without Repetition Ohio License plates have 3 #’s followed by 3 letters. EX 2: You start with 10 choices but the #s cannot be repeated. For the letters, there are 26 choices, but the letters cannot be repeated once it has been used. 6 Steps 1) # of Events: a) Event 1: Numbers b) Event 2: Letters 2) Items per Event a) Event 1: 10 Digits (can’t repeat same #) b) Event 2: 26 Letters (can’t repeat) 3) Indicate the Conditions License Plate Format (3 #’s for Event 1 NO repeats, then 3 letters for Event 2, NO repeat) 4) Set up the “blanks” needed for each Event & fill in the # of items left over. Event 1: 3 #s Event 2: 3 Letters _____ _____ _____ ______ ______ ______ 1st 2nd 3rd 1st 2nd 3rd 5) Multiply the Items: 10*9*8*26*25*24= 11,232,000 plates 6) Concl Sentence: 11,232,000 different plates can be made if the digits and letters cannot be repeated. 10 9 26 25 24 8

  4. Notes #23 Introduction to Permutations Section 11.1

  5. What is a Permutation? ANS: Permutationis an ordered arrangement of items that occurs when: • No item is used more than once. • The order of arrangement makes a difference. Section 11.2

  6. In general, the # of permutations of “n” # of item is: n! = n*(n-1)*(n-2)*… n! is also called “n”factorial

  7. Factorial Notation • The product 7∙6∙5∙4∙3∙2∙1 is called factorial and is written 7! • 0! (zero factorial), by definition, is 1. 0! = 1 Section 11.2

  8. Practice Problem #1Using Factorial Notation • Evaluate without using your calculator: ∕ ∕ ∕ ∕ Section 11.2

  9. Definition of Permutation EX 1: How many permutations/ways are possible using the following three colors: red, white, and black? Set Up: (Indicate # of choices for the 3 different colors) a) 3 choices for the first color b) 2 choices for the second color c) only 1 choice for the third color. Choices: ______ ______ ______ Colors: 1st 2nd 3rd 3 2 1 d) Can be written as 3! or 3 factorial = 3 • 2 • 1 e) Final answer = 6 permutations are possible for the 3 colors

  10. EX 2: How many different ways or permutations can the letters A, B, C, & D be arranged? You can use the Fund. Counting Principal to determine the number of permutations of n items. Set Up: (Indicate # of choices for the 4 different Letters) a) 4 choices for the first letter (from A, B, C, D) b) 3 choices for the second letter c) 2 choices for the third letter d) 1 choice for the 4th letter Choices: ______ ______ ______ ______ Letters: 1st 2nd 3rd 4th 4 1 3 2 e) Can be written as 4! or 4 factorial = 4 • 3 • 2 • 1 f) Final answer = 24 permutations/ways are possible for the letters A, B, C, & D be arranged.

  11. Definition of nPr When Order Matters use Permutations Permutations is an arrangement of outcomes in which the ORDER DOES matter Ex 3: Five projects are entered in a science contest. In how many ways can the projects come in first, second, and third? 1st 2nd 3rd Award Placement 3 projects 5 4 3 __ x __ x __ 5 projects 4 projects FINAL ANS: 5 • 4 • 3 = 60 ways

  12. Ex 4: How many different ways can 12 skiers in the Olympic finals finish the competition? (if there are no ties) Award Placement 1st 2nd 3rd 4th 5th 6th ………12th 12 11 10 9 8 7 1 _____ ________ ______ ______ _______ _______ …. _____ 12 skiers 1 skier 12! = SHOW WORK = 12*11*10*9*8*7*6*5*4*3*2*1 There are 479,001,600 different ways that 12 skiers in the Olympic finals can finish the competition.

  13. Ex 5: How many different ways can 3 of the skiers finish 1st, 2nd, & 3rd (gold, silver, bronze) Award Placement 1st 2nd 3rd 12 11 10 _____ ________ ______ 10 skier 12 skiers SHOW WORK:12 x 11 x 10 = 1320 ANS: The number of ways the skiers can win the medals is 1320.

  14. Notes #24 Permutations with Special Arrangements Section 11.1

  15. Permutations with Special Arrangements: Ex 1 Using the letters in the word ” SQUARE ", how many 6-letter arrangements, with no repetitions, are possible if the : a)  First letter is a vowel. STEPS: With “Arrangements” • Note the Given Info:6 letters in the word“SQUARE” 1stletter is a vowel (3 Vowels: U, A, E) 2) Put blanks down to represent the placement of the items Choices: ______ ______ ______ ______ ______ ______ Letters 1st 2nd 3rd 4th 5th 6th = 3 • 5! = 360 different arrangements w/o repetition are possible, given that the 1st letter is a vowel. 3 5 3 2 1 4 (6 -1) letters left

  16.   Using the letters in the word ”SQUARE", how many 6-letter arrangements, with no repetitions, are possible if the Ex 1b)  vowels and consonants alternate, beginning with a consonant.Set Up • Six spaces are needed for the 6-letter arrangements.  • Special Arrangement: Vowels and Consonants Alternate Beginning with a consonant.(S, Q, R )  Choices: ______ • _____ • ____• ____• _____• ____ Letters 1st 2nd 3rd 4th 5th 6th 3 3 2 2 1 1 = 36 Different Arrangements Vowels Section 11.1

  17. Counting Principles with Special Conditions(Phone numbers) EX 1: How many differentt 7 digit phone numbers are possible if the 1st digit cannot be a 0 or 1? 6 Steps 1) # of Events: a) Event 1: Phone Number with 7 digits 2) # of Choices/Items a) 10 Items per digit (Choices: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9) 3) Indicate the Conditions Phone # Format (7 digits, where 1st digit cannot be0or 1) 4) Set up the “blanks”for each of the 7 digits & fill-in # of choices per digit Event 1: Numbers with 7 digits, where 1st # can’t be a 0 or 1 Choices: _____ _____ _____ - ______ ______ ______ ______ Digits: 1st 2nd 3rd 4th 5th 6th 7th 5) Multiply the Items: 8*10*10*10*10*10*10= 8,000,000 6) Concl Sentence: 8,000,000 different phone numbers can be made if the 1st digit cannot be a 0 or 1. 8 10 10 10 10 10 10

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