Counting Rules

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Counting Rules
Fundament Counting Rule
In a sequence of 𝑛 events in which the first has π‘˜1 possibilities
and the second event has π‘˜2 and the third has π‘˜3 , and so on.
The total number of possibilities will be
π‘˜1 ⋅ π‘˜2 ⋅ π‘˜3 ⋅⋅⋅ π‘˜π‘›
Fundament Counting Rule
A coin is tossed and a die is rolled. Find the number of
outcomes for the sequence of events.
Fundament Counting Rule
A coin is tossed and a die is rolled. Find the number of
outcomes for the sequence of events.
2 ⋅ 6 = 12
Fundament Counting Rule
Assume that a criminal is found using your social security
number and claims that all of the digits were randomly
generated. What is the probability of getting your social
security number when randomly generating nine digits?
Fundament Counting Rule
Assume that a criminal is found using your social security
number and claims that all of the digits were randomly
generated. What is the probability of getting your social
security number when randomly generating nine digits?
9 digits each with 10 outcomes is 109 = 1,000,000,000
So
1
1,000,000,000
Fundament Counting Rule
Consider the following problem given on a history test:
Arrange the following events in chronological order.
a. Boston Tea Party
b. Teapot Dome Scandal
c. The Civil War
The correct answer is a, c, b. What is the probability of
guessing the correct order?
Factorials
The Factorial symbol (!) denotes the product
of decreasing positive whole numbers.
Ex: 4! = 4 ⋅ 3 ⋅ 2 ⋅ 1 = 24. By definition 0! = 1
Factorial Rule
A collection of 𝑛 different items can be
arranged in 𝑛! Different ways.
Factorial Rule/Permutations
During the summer, you are planning to visit
these six national parks: Glacier, Yellowstone,
Yosemite, Arches, Zion, and Grand Canyon.
You would like to find the most efficient route.
How many different routes are possible?
Permutations
Now say we only have time to visit 4 of the 50
state capitals what are the number of
different routes?
Permutations
Now say we only have time to visit 4 of the 50
state capitals what are the number of
different routes?
50!
= 50 ⋅ 49 ⋅ 48 ⋅ 47 = 5,527,200
46!
Permutations
The arrangement of 𝑛 objects in a specific
order using r objects at a time is written with
𝑛!
formula 𝑛 π‘ƒπ‘Ÿ =
𝑛−π‘Ÿ !
Permutations
In horse racing, a bet on an exacta in a race is
won by correctly selecting the horses that
finish first and second, and you must select
those two horses in the correct order. A horse
race has 20 horses, if you randomly select two
horses what are you chances of winning a bet
on an exacta?
Permutations
In horse racing, a bet on an exacta in a race is
won by correctly selecting the horses that
finish first and second, and you must select
those two horses in the correct order. A horse
race has 20 horses, if you randomly select two
horses what are you chances of winning a bet
on an exacta?
20!
20−2 !
=
20!
18!
= 20 ⋅ 19 = 380, so 1/380
Permutations
In horse racing, a bet on an exacta in a race is
won by correctly selecting the horses that
finish first and second, and you must select
those two horses in the correct order. A horse
race has 20 horses, if you randomly select two
horses what are you chances of winning a bet
on an exacta?
20!
20−2 !
=
20!
18!
= 20 ⋅ 19 = 380, so 1/380
Combinations
A selection of distinct objects without regard
to order is called a combination.
Combinations
A selection of distinct objects without regard
to order is called a combination.
Given the letters A, B, C, and D, list the
permutations and combinations for selecting
two letters.
Combinations
The number of combinations of π‘Ÿ objects
selected from n objects is denoted by 𝑛 πΆπ‘Ÿ or
𝑛
𝑛!
=
𝑛−π‘Ÿ !π‘Ÿ!
π‘Ÿ
Combinations
How many combinations of 4 objects taken 2
at a time?
4!
4!
4⋅3
4
=
=
= =6
4−2 !2!
2!2!
2⋅1
2
Combinations
The Foreign Language Club is showing a fourmovie marathon of subtitled movies. How
many ways can they choose 4 from the 11
available?
Combinations
The Foreign Language Club is showing a fourmovie marathon of subtitled movies. How
many ways can they choose 4 from the 11
available?
11!
11⋅10⋅9⋅8⋅7⋅6 ⋅5
11
=
=
=15
11−4 !4!
4⋅3⋅2⋅1
4
Combinations
How many ways can a jury of 6 women and 6
men be selected from 10 women and 12 men?
Combinations
How many ways can a jury of 6 women and 6
men be selected from 10 women and 12 men?
10
12
⋅
= 16 ⋅ 18 = 224
6
6
Homework!
4-4: 1-59 odd
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