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pencil, highlighter, GP notebook, textbook, calculator

total:. pencil, highlighter, GP notebook, textbook, calculator. U7D1. Have out:. Bellwork:. Eugene needs to create a 5 digit code for his home’s security system. He can choose numbers from 0 – 9. (Hint: how many numbers are there from 0 – 9? Use your hands to count it out.).

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pencil, highlighter, GP notebook, textbook, calculator

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  1. total: pencil, highlighter, GP notebook, textbook, calculator U7D1 Have out: Bellwork: Eugene needs to create a 5 digit code for his home’s security system. He can choose numbers from 0 – 9. (Hint: how many numbers are there from 0 – 9? Use your hands to count it out.) a) If repeated numbers are allowed, how many codes are possible? b) If repeated numbers are NOT allowed, how many codes are possible?

  2. total: Eugene needs to create a 5 digit code for his home’s security system. He can choose numbers from 0 – 9. (Hint: how many numbers are there from 0 – 9? Use your hands to count it out.) a) If repeated numbers are allowed, how many codes are possible? +2 +2     = 100000 10 10 10 10 10 1st digit 2nd digit 3rd digit 4th digit 5th digit b) If repeated numbers are NOT allowed, how many codes are possible? +2 +2     = 30240 10 9 8 7 6 1st digit 2nd digit 3rd digit 4th digit 5th digit

  3. Factorials & The Fundamental Counting Principle Take out the worksheet • California license plates are seven digits long, in the form # L L L # # # , where # is a digit from 0 thru 9, and L is one of the 26 letters of the alphabet. a) How many different license plates are possible if repeated numbers and letters are allowed?       10 26 26 26 104● 263 10 10 10 = 175,760,000 ___ ___ ___ ___ ___ ___ ___ = _________ b) How many different license plates are possible if repeated numbers and letters are NOT allowed?       10 26 25 24 9 8 7 78,624,000 ___ ___ ___ ___ ___ ___ ___ = _________

  4. Factorials n! is pronounced “n factorial” n! __________= descending consecutive This is a shorthand way for multiplying ____________________ whole numbers. In other words, n! is the product of all positive integers less than or equal to n. Example: Simplify 4! 4! = 4  3 2 1 = 24 Your graphing calculators can also compute 4!. Type: MATH  PROB 4: !

  5. Factorials Practice: Expand and evaluate the following: a) 6! = 6  5  4  3 2 1 = 720 b) 2! = 2  1 = 2 c) 10! = 10  9  8  7 6  5  4  3 2 1 = 3,628,800

  6. For the next problem, we need 5 volunteers to help out.

  7. 2. Five friends go to the movies. In how many ways can they arrange themselves if: a) There are no restrictions?     4 5! = 120 5 3 2 1 _____ _____ _____ _____ _____ = _____

  8. Frasers Bennifer Brangelina Continuing with the 5 friends at the movies, determine the following scenarios: b) 2 couples and 1 loner: _____ _____ _____ _____ _____ = ____________ We still have 5 people, however, when it comes to couples, they will want to sit next to their partners. Have you ever met couples who are always together that we might as well just consider them “one” person? So, when we solve this problem, treat couples as “one.” Keep in mind that we need to consider how they will arrangement themselves next to each other.

  9. Continuing with the 5 friends at the movies, determine the following scenarios: b) 2 couples and 1 loner: 3  2  1 3! _____ _____ _____ _____ _____ = ____________ _____ _____ _____ = _____ Consider: how many groups are we arranging? 3 groups: 2 couples and 1 loner C1 C2 L We are not quite done yet… The couples can also be arranged between partners. ( ) ( ) 2  1 2  1 _____ _____ _____ _____ _____ ♥ C2 ♥ ♥ C1 ♥ L

  10. Continuing with the 5 friends at the movies, determine the following scenarios: b) 2 couples and 1 loner: 3  2  1 3! _____ _____ _____ = _____ ( ) ( ) 2  1 2  1 _____ _____ _____ _____ _____ ♥ C2 ♥ ♥ C1 ♥ L Now we have exhausted all possibilities. Let’s put it all together. 3!  2!  2! = 24 # of ways to arrange 3 groups # of ways to arrange Couple #1 # of ways to arrange Couple #2

  11. Continuing with the 5 friends at the movies, determine the following scenarios: c) 1 couple and 3 others: _____ _____ _____ _____ _____ = ____________ Just like the previous problem, consider: 4 groups: How many “groups” are we arranging? C L1 L3 L2 4  3  2  1 = 4! Also, how many ways can the couple arrange themselves? ( ) 2  1 _____ _____ _____ _____ _____ L1 L3 L2 ♥ C1 ♥ Put it all together: 4!  2! = 48

  12. Continuing with the 5 friends at the movies, determine the following scenarios: d) 1 set of conjoined twins (attached at the hip) with 3 others: _____ _____ _____ _____ _____ = ____________ 4 groups: How many “groups” are we arranging? T L1 L3 L2 4  3  2  1 = 4! Are there other ways to arrange the conjoined twins? No! Put it all together: 4! = 24

  13. Continuing with the 5 friends at the movies, determine the following scenarios: e) 1 set of conjoined twins (attached at the hip), their dates, and 1 loner: Date #2 Date #1 Date #1 Date #2 ( twins ) ( twins ) loner loner _____ _____ _____ _____ _____ = ____________ Where can the twins sit so that their dates will be next to them? Where will the loner sit? How many different arrangements are possible? 2! = 2

  14. You are selecting classes for next year. You plan to take Math Analysis, AP U.S. History, AP English, AP Biology, AP Spanish, and AP Psychology. IC – 17 a) How many schedules are possible?      = 6! = 720 6 5 4 3 2 1 4th period 6th period 3rd period 5th period 2nd period 1st period b) Assuming all classes are independent, what is the probability of getting 1st period Math Analysis? P (1st period Math Analysis) = c) What is the probability of 1st period Math Analysis and 2nd period AP Biology? P (1st period MA, 2nd period Bio) = 

  15. IC – 20 How many distinguishable batting orders can be made from the nine starting players on a baseball team?         9 8 7 6 5 4 3 2 1 4th player 6th player 3rd player 5th player 7th player 9th player 8th player 2nd player 1st player = 9! “Distinguishable” implies that once a player is given a position, they cannot fill another spot. = 362880 How many distinct 9 – letter “words” are possible from the letters in the word FRACTIONS? IC – 21         9 8 7 6 5 4 3 2 1 4th letter 6th letter 3rd letter 5th letter 7th letter 9th letter 8th letter 2nd letter 1st letter “Distinct” implies that no “words” are alike. = 9! = 362880

  16. a) How many distinct words are possible from the word MOON? EXTENSION: We may be tempted just to compute:    4 3 2 1 = 4! = 24 Let’s list the possibilities: Are all the words distinct from one another? MNOO NMOO OMNO ONOM No! MNOO NMOO OMNO ONOM Notice that all the words are repeated. MONO NOMO OMON OOMN Why? MONO NOMO OMON OOMN There is no distinction between the O’s. MOON NOOM ONMO OONM MOON NOOM ONMO OONM How can we eliminate this double counting? We must divide by 2. More specifically, we will divide by 2!. 4! Now, we are just counting the distinct words, and we have eliminated repeats. = 12 2!

  17. EXTENSION: b) How many distinct 6 – letter “words” are possible from the letters in the word ROEDER? What letters are repeated? R and E Be careful… both of these letters will double count all the words. Since there are 2 sets of repeating letters, we need to divide by 2! for each set. 6! = 180 2! 2! Try parts (b) and (c) on your own.

  18. EXTENSION: c) How many distinct 7 – letter “words” are possible from the letters in the word SELLERS? 2–S’s, 2–L’s, 2–E’s 7! = 630 2! 2! 2! d) How many distinct 11 – letter “words” are possible from the letters in the word MISSISSIPPI? 4–I’s, 4–S’s, 2–P’s 11! = 34650 4! 2! 4!

  19. You don’t say! IC – 23 Remembering what n! means can help you do some messy calculations quickly, as well as help you do problems that might be too large for your calculator. For instance, use your calculator to compute . What happened? OVERFLOW ERROR Believe it or not, we can compute without a calculator. Let’s try a nicer example first. 9  8  7  6  5  4  3  2  1 Example: = 9  8  7 = 504 6  5  4  3  2  1

  20. IC – 23 6! 9  8  7  6  5  4  3  2  1 = 9  8  7 = 504 6  5  4  3  2  1 This method isn’t too bad, but is there a faster method? Notice that there is a 6! “buried” within 9!. Expand 9! until you reach 6!. Try the example again. 9  8  7  6! = 9  8  7 = 504 6! Try parts (a) through (d) on your own.

  21. IC – 23 10  9  8! a) = 10  9 = 90 8! 20  19  18! b) = 190 18! 2! 7  6  5  4! c) = 35 4! 3! 75  74  73! d) = 75  74 = 5550 73!

  22. Finish the worksheet and IC 24 – 29, 31, 32

  23. I’m kinda bored. I’ll save you from Jason. Jason ain’t got nothin’ on me.

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