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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
b c
c ba c
c aa b
b a
Aim: Permutations Course: Math Lit..
Counting Principle at Work
6 • 5 • 4 • 3 • 2 • 1 = 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?
9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 = 362,880
w/o repetition
Tom, Dick, Harry, Mary Larry & Joe
Factorial: n! = n•(n - 1)(n - 2)(n - 3). . . . 1. When arranging all objects in the set
Aim: Permutations Course: Math Lit..
Counting Principle at Work
6 • 5 • 4 = 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?
9 • 8 • 7 • 6 • 5 • 4 = 60480
Since our selection is limited
to only 3 names we stop after the
3rd selection
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’? 5 • 4 • 3 = 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
nPn = n! = n(n – 1)(n – 2)(n – 3) . . . 3, 2, 1
The number of permutation of n things taken r at a time is
( 1)( 2)n rP n n n L
r factors
!
( )!
n
n r
Factorial: n! = n•(n - 1)(n - 2)(n - 3). . . . 1. 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
5P5
8P6
6P1
= n! = 5 • 4 • 3 • 2 • 1 = 120
n! (n - r)!
= 8! (8 - 6)!
=
8 • 7 • 6 • 5 • 4 • 3 • 2 • 1
2 • 1
8! 2!
=
= 20160
6! 5!
= = 6
3P3 = 6
4P3 = 24
5P3 = 60
Aim: Permutations Course: Math Lit..
Model Problems
How many arrangements can be made from the word ANGLE?
5P5 = 120
How many arrangements can be made from the word ANGLE if repetition of letters is allowed?
5 • 5 • 5 • 5 • 5 = 55 = 3125
Repetition is key issueRepetition 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
7 • 6 • 5 • 2 • 1 = 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 EATATETAEETAAET
TEE EET ETE TEE ETE EET
Duplicates
6
Repeated letters alter number of arrangements possible because they lead to duplicated ‘words’.
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!
2!= 3
!
!
n
r !n nP
r
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
n!
r!s! t!
s = 4, the number of times S is repeated
s = 2, the number of times P is repeated
11!
4!4!2! = 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
n!
r!s! t!
5!
2! = 60
= 205!
3!
= 305!
2!2!
5! = 120
= 105!
3!2!
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 E1 E2 E3 E4 E5
Permutations
9 letters altogether; A and E are only vowels; 7 consonants for 1st three events
7 6 5 2 1· · · · = 420
)!(
!
rn
nPrn
)!(
!
rn
nPrn
7 3P
210
2 2P
2
7!
(7 3)!2! = 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? 10P3
How messages are possible if the top flag must be one of three specific flags? 3 • 9 • 8 = 216
How many different ways can John place a math book, a history book, a science book and an art book on a shelf? 4P4 = 4! = 24
How many different ways can John place the 4 books if the math book must be first?
1 • 3 • 2 = 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
)!(
!
rn
nPrn
)!(
!
rn
nPrn
7 P4 7!
(7 4)!
7!
3!
7 6 5 4 3 2 1
3 2 1
7 6 5 4 840
AFTER THE A IS PLACED, THERE ARE 6 LETTERS LEFT FOR 3 PLACES OR 6P3
1 • = 1206P3
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
6 5 1 4• • • = 120
ONLY 1 CHOICE
FOR THIRD LETTER - A
AFTER THE A IS PLACED, THERE ARE 6 LETTERS LEFT FOR 3
PLACES OR 6P3
1 • 6P3 = 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 5 4 2 1• • • = 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 5P2
5P2 = 402 1• •
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
1 5 4 3 2 1• • • • •
ONLY 1 CHOICE FOR FIRST LETTER - S
5P5
= 120AFTER THE S IS PLACED, THERE ARE 5 LETTERS LEFT FOR 5 PLACES OR 5P5
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
3 2 13 2 1• • • • • = 36
VOWELS - U, A, E CONSON. - S, Q R
ONLY 3 CHOICES
U,A,E
ONLY 3 CHOICES
S,Q,R
• = 363P33P3
Aim: Permutations Course: Math Lit..
Model Problems
AMDKDLK3EW
AMDKDLK3EW
AMDKDLK3EW
AMDKDLK3EW
AMDKDLK3EW
AMDKDLK3EW
AMDKDLK3EW
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?
2 1 4 3 2 1• • • • • = 1443 •
First three books are chemistry
Last four books are
biology
3P3 4P4
3P3 • 4P4 = 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!
r!s! t!n = # bulbs = 46r = # of tulips = 8s = # of daffodils = 10
t = # of crocus = 28
46!
8!10!28!= 1.233. . . x 1017
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 60 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 • = 60ONLY 3 CHOICES FOR FIRST DIGIT 7,8,9
5 4•
5P2
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.