Chapter 7: Section 7-2
Basic Counting Principles (Revisited)
D. S. Malik
Creighton University, Omaha, NE
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Multiplication Principle
Theorem
( Multiplication Principle) Suppose that a multistep experiment consists
of k (independent) steps. Suppose that the …rst step can be completed in
n1 ways, the second step can be completed in n2 ways regardless of the
results of step 1, the third step can be completed in n3 ways regardless of
the results of steps 1 and 2, and so on. In general, suppose that step i,
can be completed in ni ways, regardless of the results of the …rst i 1
steps. Then the entire experiment can be completed in
n1 n2
nk
ways.
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Example
Suppose we want to …nd the number of words of length 3 that can be
written using the letters A, B, C , D, and E such that no repetition of
letters in a word is allowed. The task of forming a three letter word can be
thought of as a three step experiment. In the …rst step, we select the …rst
word. This can be done in 5 ways because we have 5 letters to choose
from. In the second step, we select the second letter. Because the
repetition of letters is not allowed and we have already used one letter in
step 1, the second step can be completed in 4 ways. Similarly, the third
step can be completed in 3 ways. Hence, by the multiplication principle,
the number words of length 3 is
5 4 3 = 60.
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Example
Find the number of license plates that can be written with three uppercase
letters followed by three digits.
There are 26 uppercase letters and 10 digits. Now a license plate consists
of 3 letters followed by 3 digits:
l1 l2 l3 d1 d2 d3
This counting problem can be viewed as a task consisting of 6 successive
steps. In each of the steps 1, 2, and 3, choose an uppercase letter, in
each of the steps 4, 5,and 6, choose a digit. Because there are 26 letters,
each of the steps 1, 2, and 3 can be completed in 26 ways. Similarly,
because there are 10 digits, each of the steps 4, 5, and 6 can be completed
in 10 di¤erent ways. Thus, a license plate can be formed in:
26 26 26 10 10 10 = 17, 576, 000
ways. Hence, there are 17, 576, 000 di¤erent license plates with three
uppercase letters followed by three digits.
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Addition Principle
Let X be a set such that n (X ) = m1 and Y be a set such that
n (Y ) = m2 . If X \Y = ∅, then n (X [ Y ) = m1 + m2 .
Theorem
Let X1 , X2 , . . . , Xk be sets such that the number of elements in Xi is mi ,
that is, n (Xi ) = mi , i = 1, 2, . . . , k, k 2. Suppose that for any two sets
Xi and Xj , Xi \ Xj = ∅, i = 1, 2, . . . , k, j = 1, 2, . . . , k, i 6= j. That is, the
sets X1 , X2 , . . . , Xk are pairwise disjoint. Then
n ( X1 [ X2 [
[ Xk ) = m1 + m2 +
+ mk .
De…nition
Suppose a task T is a collection of a sequence of tasks T1 , T2 , . . . , Tk .
The tasks T1 , T2 , . . . , Tk are called independent if the outcome of any
task, say Ti , does not in‡uence the outcome of any other task in the
collection or sequence.
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Addition Principle
Addition Principle: Suppose that the tasks T1 , T2 , . . . , Tk can be done
in n1 , n2 , . . . , nk ways, respectively. If all these tasks are
independent of each other, then the number of ways to do
one of these tasks is n1 + n2 +
+ nk .
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Example
There are three boxes containing books. The …rst box contains 15
mathematics books by di¤erent authors, the second box contains 12
chemistry books by di¤erent authors, and the third box contains 10
computer science books by di¤erent authors. A student wants to take a
book from one of the three boxes. Suppose tasks T1 , T2 , and T3 are as
follows:
T1 : choose a mathematics book.
T2 : choose a chemistry book.
T3 : choose a computer science book.
Then tasks T1 , T2 , T3 can be done in 15, 12, and 10 ways, respectively. All
these tasks are independent of each other. Hence, the number of ways to
do one of these tasks is 15 + 12 + 10 = 37.
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Example
For breakfast the following items are available: 4 types of cereal, 2 types
of juice, and 3 types of bread. Let us determine the number of ways a
breakfast can be prepared if exactly two items are selected from two
di¤erent groups.
Because exactly two items must be selected from two di¤erent groups, a
breakfast can be prepared in either of the following three ways: (i) a cereal
and a juice, or (ii) a cereal and a bread, or (iii) a juice and a bread.
Let
X be the set of breakfasts that consist of a cereal and a juice;
Y be the set of breakfasts that consist of a cereal and a bread; and
Z be the set of breakfasts that consist of a juice and a bread.
Then X [ Y [ Z is the set of all breakfasts that consist of exactly two
items from two di¤erent groups.
Notice that X \ Y = ∅, X \ Z = ∅, and Y \ Z = ∅, that is, the sets
X , Y , and Z are pairwise disjoint.
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4 types of cereal, 2 types of juice, and 3 types of bread. Example
Continued.
Example
A breakfast in the set X consists of a cereal and a juice. Thus, we can
…rst select a cereal and then a juice. Now a cereal can be selected in 4
ways and a juice can be selected in 2 ways. Thus, by the multiplication
principle, a breakfast consisting of a cereal and a juice can be prepared in
4 2 = 8 ways. Therefore, n (X ) = 8. In a similar manner,
n (Y ) = 4 3 = 12 and n (Z ) = 2 3 = 6. Consequently,
n (X [ Y [ Z ) = n (X ) + n (Y ) + n (Z ) = 8 + 12 + 6 = 26.
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Exercise: How many license plates consisting of 3 letters followed by 4
digits can be prepared if repetitions are not allowed?
Solution: A licence plate with 3 letters followed by 4 digits is of the
form
l1 l2 l3 d1 d2 d3 d4 ,
where li represents a letter and di represents a digit. Because
repetitions are not allowed, there are 26 choices for l1 , 25
choices for l2 , 24 choices for l3 , 10 choices for d1 , 9 choices
for d2 , 8 choices for d3 , and 7 choices for d4 . Hence, by the
multiplication principle, the number of required licence plates
is
26 25 24 10 9 8 7 = 78, 624, 000.
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Exercise: Suppose there is a set of 4 distinct mystery novels, 5 distinct
romance novels, and 3 distinct poetry books.
(a) In how many ways can these books can be arranged on a
shelf?
(b) In how many ways can these books be arranged on a
shelf if all poetry books stay together?
Solution: (a) There are a total of 4 + 5 + 3 = 12 books. Thus, the
number of ways these books can be arranged is
12! = 479, 001, 600.
(b) The 3 poetry books and these must stay together. Now,
3 poetry books can be arranged in 3! = 6 ways.
Now, there are 4 mystery novels and 5 romance novels. The
required arrangement can be viewed as arranging 10 books,
3 poetry books together, 4 mystery novels, and 5 romance
novels. Hence, the number of such arrangements is
10! 3! = 21, 772, 800.
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Exercise: To attract customers, a new car dealer advertised a summer
sale with special discounts and …nances. The dealer has 7
di¤erent models available. Each model comes in 5 di¤erent
colors, and has a choice of leather or cloth interior. The
customer can choose from 2 di¤erent music systems.
Moreover, each model has 3 di¤erent exterior designs. Find
the number of ways a customer can buy a car.
Solution: There are 7 choices for the models, 5 choices for colors, 2
choices for the fabric, 2 choices for the music system, and 3
choices for the exterior design. Hence, the total number of
choices is 7 5 2 2 3 = 420.
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