pencil highlighter GP notebook textbook calculator Eugene needs

pencil, highlighter, GP notebook, textbook, calculator 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? 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 +2 +2 possible? 10 1 st digit 10 2 nd digit 10 3 rd digit 10 4 th digit 10 = 100000 5 th digit b) If repeated numbers are NOT allowed, how many codes are possible? +2 +2 10 1 st digit 9 2 nd digit 8 3 rd digit 7 4 th digit 6 = 30240 5 th digit total:

1. 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 ___ 104 ● 263 = 175, 760, 000 10 ___ 10 = _____ ___ 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 ___

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: !

Expand 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

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

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 _____ = _____

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. Frasers to consider how they will arrangement themselves next to each other. Bennife Brangelina r

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 C 1 C 2 L We are not quite done yet… The couples can also be arranged between partners. 2 _____ 1 2 _____ 1 _____ ( )( ♥ C 1 ♥ ) ♥ C 2 ♥ L

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 )_____ ♥ C 2 ♥ ♥ C 1 ♥ L Now we have exhausted all possibilities. Let’s put it all together. 3! 2! = 24 # of ways to arrange 3 groups Couple #1 Couple #2

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

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: _____ _____ = ______ How many “groups” are we arranging? 4 groups: T L 1 L 2 L 3 4 3 2 1 = 4! Are there other ways to arrange the conjoined twins? No! Put it all together: 4! = 24

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 loner ( twins ) #2 loner #1 ( twins ) #1 #2 _____ _____ = ______ 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

IC – 17 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. a) How many schedules are possible? 3 2 6 5 4 1 = 6! = 720 4 th 6 th 3 rd 5 th 2 nd 1 st period period b) Assuming all classes are independent, what is the probability of getting 1 st period Math Analysis? P (1 st period Math Analysis) = c) What is the probability of 1 st period Math Analysis and 2 nd period AP Biology? P (1 st period MA, 2 nd period Bio) =

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

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

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.

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! 4! 4! 2! = 34650

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. You don’t say! Let’s try a nicer example first. Example: 9 8 7 6 5 4 3 2 1 = 9 8 7 = 504

IC – 23 6! 9 8 7 6 5 4 3 2 1 = 9 8 7 = 504 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! 6! = 9 8 7 = 504 Try parts (a) through (d) on your own.