Math 160 - Cooley
Intro to Statistics
OCC
Section 4.8 – Counting Rules
The Basic Counting Rule (BCR)
Suppose that r actions are to be performed in a definite order. Further suppose that there are m1 possibilities for
the first action and that corresponding to each of these possibilities are m2 possibilities for the second action, and
so on. Then there are m1 · m2 · · · mr possibilities altogether for the r actions.
Factorials
The product of the first n positive integers (counting numbers) is called n factorial and is denoted n!, where
n!  n(n  1)(n  2)    (3)(2)(1)
We also define 0! = 1
The Permutations Rule
The number of possible permutations (ordered arrangements) of r objects from a collection of m objects is
given by the formula:
m!
mPr =
(m  r )!
The Special Permutations Rule
The number of possible permutations (ordered arrangements) of m objects among themselves is m!
The Combinations Rule
The number of possible permutations (unordered arrangements) of r objects from a collection of m objects is
given by the formula
m
m!
mCr  

 r  r !( m  r )!
 Exercises:
1)
In the U.S., a five-digit zip code consists of five digits, of which the first three give the sectional center
and the last two the post office or delivery area. In addition to the five-digit zip code, there is a trailing
plus four zip code. The first two digits of the plus four zip code give the sector or several blocks and the
last two the segment or side of the street.
Suppose for the five-digit zip code, the first four digits can be any of the digits 0–9 and the fifth any of
the digits 1–8. For the plus four zip code, the first three digits can be any of the digits 0–9 and the fourth
any of the digits 1–9.
a)
How many possible five-digit zip codes are there?
b)
How many possible plus four zip codes are there?
c)
How many possibilities are there in all, including both the five-digit zip code and the plus four
zip code?
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Math 160 - Cooley
Intro to Statistics
OCC
Section 4.8 – Counting Rules
 Exercises:
2)
Heidi has 6 necklaces and 8 pairs of earrings. In how many ways can she select a necklace and a
pair of earrings to wear?
3)
Alfred is opening a new office and needs to decorate and furnish it. He has 3 sources for
wallpaper, 5 sources for carpet, 4 sources for drapes, 8 sources for furniture, and 2 sources for
pictures. How many ways can he select one of each?
4)
5!
8)
In how many ways can 8 CD’s be arranged on a shelf?
9)
In how many ways can a sorority of 20 members select a president, vice president and treasury,
assuming that the same person cannot hold more than one office.
10)
5C3
12)
In how many ways can a sorority of 20 members select three members to serve on a committee?
13)
A boss has 8 employees and 5 are chosen to give a presentation. How many different ways can the
boss choose the presenters if the order of the presenters is not important?
14)
A key pad lock has 10 different digits, and a sequence of 4 digits must be selected for the lock to open.
How many key pad codes are possible?
15)
A key pad lock has 10 different digits, and a sequence of 4 different digits must be selected for the
lock to open. How many key pad codes are possible?
5)
11)
8!
6)
5P2
7)
6P3
8C2
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Math 160 - Cooley
Intro to Statistics
OCC
Section 4.8 – Counting Rules
 Exercises:
16)
A textbook search committee is considering seven books for possible adoption. The committee has
decided to select three of the seven for further consideration. In how many ways can they do so?
17)
If a softball league has 10 teams, how many different end of the season rankings are possible?
(Assume no ties).
18)
Five Card Draw. A hand of five-card draw poker consists of an unordered arrangement of five cards from a
standard deck of 52 cards.
19)
a)
How many five-card draw poker hands are possible?
b)
How many different hands consisting of two Aces, two 8’s, and a Jack are possible?
c)
The hand in part a is called “Two Pair”: two cards of one denomination, two cards of a second
denomination, and one card of a third denomination. How many different hands consisting of
two pairs are possible?
d)
A “Four of a Kind” is where there are four cards of one denomination. How many different
hands consisting of a four of a kind are possible?
In 1988, the California Lottery was initiated. There were 49 lotto balls, each numbered from 1 to 49.
In order to play, you must choose 6 numbers.
a)
In how many ways can someone choose 6 numbers from the 49?
b)
A year later, they added 4 more lotto balls, to make a total of 53 lotto balls, each numbered from
1 to 53. In how many ways can someone choose 6 numbers from the 53?
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Math 160 - Cooley
Intro to Statistics
OCC
Section 4.8 – Counting Rules
 Exercises:
20)
The Birthday Problem (The Birthday Paradox). A math class has 32 students. Find the probability
that at least 2 students in the class have the same birthday. For simplicity, assume that there are always
365 days in a year and birth rates are constant throughout the year.
Notes:
 365   364  
 P
P(that 2 students do share a birthday)  1  

 1   365 22   0.00274



 365 
 365   365  
 365   364   363  
 P
P(that 3 students do share a birthday)  1  


 1   365 33   0.008204




 365 
 365   365   365  
 365   364   363   362  
 P
P(that 4 students do share a birthday)  1  



 1   365 44   0.016356





 365 
 365   365   365   365  
 365   364   363 
P
 365  ( n  1)  
P(that n students do share a birthday)  1  


 
 1  365 nn





365
365


 365   365   365 
Note: For values of n  40 , the calculator will create an error, due to overflow, since 365 P40  110100 . Thus, for
values n  40 , a much more powerful calculator or computer must be used for computations.
Answer:
P(that 32 students do share a birthday)  1 
P
 0.7533
365
365 32
32
http://www.cornell.edu/video/the-tonight-show-with-johnny-carson-feb-6-1980-excerpt
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