Aim: Permutations Course: Math Lit.. Aim: Is Spiderman a permutation or what? A = {a, b, c, d, e} What could be a two element subset of A? {a, b} Is = {b, a} ? Yes order of elements in set is not important

Aim: Permutations Course: Math Lit.. Subsets & Arrangements If order were important {a, b} is = {b, a} ? No A = {a, b, c, d, e} If the two elements a and b are selected from A, then there is one subset (order not important): {a, b} there are two arrangements (order important): (a, b) and (b, a) note: both are selected without repetitions (without replacement)

Aim: Permutations Course: Math Lit.. Model Problem Consider selecting the elements a, b, and c from set A = {a, b, c, d, e}. List all possible subsets of these elements, as well as all possible arrangements. 1 subset: {a, b, c} 6 arrangements: (a, b, c), (a, c, b), (b, a, c), (b, c, a), (c, a, b), (c, b, a) a b c bc cb ac ca ab ba

Aim: Permutations Course: Math Lit.. Counting Principle at Work = 720 How many different arrangements can be made using all six names? 5 ways to choose 2nd name 4 way to choose 3rd name 6 ways to choose 1st name 2 ways to choose 5th name 1 way to choose 6th name 3 ways to choose 4th name How many different arrangements can be made using 9 names? = 362,880 w/o repetition Tom, Dick, Harry, Mary Larry & Joe Factorial: n! = n(n - 1)(n - 2)(n - 3) When arranging all objects in the set

Aim: Permutations Course: Math Lit.. Counting Principle at Work = 120 How many different arrangements using only 3 names can be made from six names? 5 ways to choose 2nd name 4 ways to choose 3rd name 6 ways to choose 1st name How many different arrangements using 6 names from a selection of 9 names? = Since our selection is limited to only 3 names we stop after the 3rd selection How many three letter ‘words’ can be made from the word ‘STUDY’? = 60 Tom, Dick, Harry, Mary Larry & Joe

Aim: Permutations Course: Math Lit.. Permutations A permutation is an arrangement of objects in a specific order. The number of permutation of n things taken n at a time is n P n = n! = n(n – 1)(n – 2)(n – 3)... 3, 2, 1 The number of permutation of n things taken r at a time is r factors Factorial: n! = n(n - 1)(n - 2)(n - 3) When arranging all objects in the set: Definition - 0! = 1.

Aim: Permutations Course: Math Lit.. Model Problems How many three letter ‘words’ can be made from the letters of CAT GOAT CLONE 5P55P5 8P68P6 6P16P1 = n! = = 120 n! (n - r)! = 8! (8 - 6)! = ! 2! = = ! 5! = = 6 3P33P3 4P34P3 = 24 5P35P3 = 60

Aim: Permutations Course: Math Lit.. Model Problems How many arrangements can be made from the word ANGLE? 5 P 5 = 120 How many arrangements can be made from the word ANGLE if repetition of letters is allowed? = 5 5 = 3125 Repetition is key issue How many 5-letter arrangements can be made from E, Q, B, X, R, T, L, A, and V if the last two letters must be vowels. no repetition = 420

Aim: Permutations Course: Math Lit.. Repetitions How many 3-letter arrangements can you make from the word TEA? How many 3-letter arrangements can you make from the word TEE? TEA EATA TETA EETA AET TEE EET ETE TEE ETE EET Duplicates 6 Repeated letters alter number of arrangements possible because they lead to duplicated ‘words’. 3

Aim: Permutations Course: Math Lit.. Permutations with Repetitions The number of permutations of n things, taken n at a time with r of these things identical, is given by How many 3-letter arrangements can you make from the word TEE? n = 3 letters to arrange r = 2, the number of times E is repeated = 3

Aim: Permutations Course: Math Lit.. Permutations with Repetitions The number of permutations of n things, taken n at a time with r things identical, s things identical, and t things identical, is given by Find the number of permutations of the letters in MISSISSIPPI. n = 11 letters to arrange r = 4, the number of times I is repeated s = 4, the number of times S is repeated s = 2, the number of times P is repeated = 34,650

Aim: Permutations Course: Math Lit.. Model Problems How many different 5-letter permutations are there is the letters in APPLE ADDED VIVID How many different 5-digit numerals can be written using all 5 digits listed? 1, 2, 3, 4, 5 1, 1, 2, 2, 2 = 60 = 20 = 30 5! = 120 = 10

Aim: Permutations Course: Math Lit.. Model Problems How many 5-letter arrangements can be made from E, Q, B, X, R, T, L, A, and V if the last two letters must be vowels. no repetition Counting Principle E1E2 E3E4E5 Permutations 9 letters altogether; A and E are only vowels;  7 consonants for 1 st three events · · = 420 · · ·

Aim: Permutations Course: Math Lit.. Model Problems Sea-going vessels use different flags or arrangements of flags to send messages to other vessels. If 10 different flags are available, and if every message consists of three different flags, how many different messages are possible? 10 P 3 How messages are possible if the top flag must be one of three specific flags? = 216 How many different ways can John place a math book, a history book, a science book and an art book on a shelf? 4P44P4 = 4! = 24 How many different ways can John place the 4 books if the math book must be first? = 6

Aim: Permutations Course: Math Lit.. Model Problems Tell how many four letter ‘words’ can be made from the letters X, B, T, L, R, V, and A for each situation. A. There are no restrictions B. The first letter must be A ONLY 1 CHOICE FOR FIRST LETTER - A AFTER THE A IS PLACED, THERE ARE 6 LETTERS LEFT FOR 3 PLACES OR 6 P 3 1 = 120 6P36P3

Aim: Permutations Course: Math Lit.. Model Problems Tell how many four letter ‘words’ can be made from the letters X, B, T, L, R, V, and A for each situation. C. The third letter must be A = 120 ONLY 1 CHOICE FOR THIRD LETTER - A AFTER THE A IS PLACED, THERE ARE 6 LETTERS LEFT FOR 3 PLACES OR 6 P 3 1 6P36P3 = 120

Aim: Permutations Course: Math Lit.. Model Problems Tell how many four letter ‘words’ can be made from the letters X, B, T, L, R, V, and A for each situation. D.. The last two letters must be R or T in either order = 40 ONLY 2 CHOICES FOR THIRD LETTER & FOURTH LETTER AFTER THE R AND T ARE PLACED, THERE ARE 5 LETTERS LEFT FOR 2 PLACES OR 5 P 2 5P25P2 =

Aim: Permutations Course: Math Lit.. Model Problem Using the word SQUARE, find how many 6-letter arrangements with no repetitions are possible if the: A. First letter is S ONLY 1 CHOICE FOR FIRST LETTER - S 5P55P5 = 120 AFTER THE S IS PLACED, THERE ARE 5 LETTERS LEFT FOR 5 PLACES OR 5 P 5

Aim: Permutations Course: Math Lit.. Model Problem Using the word SQUARE, find how many 6-letter arrangements with no repetitions are possible if the: B. Vowels and consonants alternate beginning with a vowel = 36 VOWELS - U, A, E CONSON. - S, Q R ONLY 3 CHOICES U,A,E ONLY 3 CHOICES S,Q,R = 36 3P33P3 3P33P3

Aim: Permutations Course: Math Lit.. Model Problems AMDKD LK3EW Four different biology books and three different chemistry books are to be placed on a shelf with the biology books together and to the right of the chemistry books. In how many ways can this be done? = First three books are chemistry Last four books are biology 3P33P34P44P4 3P33P3 4P44P4 = 144

Aim: Permutations Course: Math Lit.. Model Problem Frances has 8 tulip bulbs, 10 daffodil bulbs, and 28 crocus bulbs. In how many different ways can Frances plant these bulbs in a row in her garden? n = # bulbs = 46 r = # of tulips = 8 s = # of daffodils = 10 t = # of crocus = 28 = x 10 17

Aim: Permutations Course: Math Lit.. Model Problems A combination lock has 60 numbers around its dial. How many different arrangments are possible if every combination has three numbers? 60 = 216,000 Question: can a number be used more than once? A three digit number is formed by selecting from the digits 4, 5, 6, 7, 8, and 9 with no repetitions. How many of these 3-digit numbers will be greater than 700? 3 = 60 ONLY 3 CHOICES FOR FIRST DIGIT 7,8,9 54 5P25P2

Aim: Permutations Course: Math Lit.. Model Problem Find the total number of different twelve- letter arrangements that can be formed using the letters in the word PENNSYLVANIA.

Aim: Permutations Course: Math Lit..

Aim: Permutations Course: Math Lit..