discrete mathematics, 1st edition kevin ferland chapter 6 basic counting 1

Discrete Mathematics, 1st EditionKevin Ferland

Chapter 6

Basic Counting

1

6.1 The Multiplication Principle

(Dinner Choices). A guest at a formal dinner has 4 entrée choices and 2 dessert choices. If a guest’s dinner is entirely determined by these two choices, then how many different dinner choices are there?

The total number of dinner choices is given by the product 4 · 2 = 8.

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EXAMPLE 6.1 (Solution) (Cont.)

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THEOREM 6.1

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6.2 Permutations and Combinations

PermutationsA permutation of a set of objects is an

ordering of those objects. Permutations reflect selections for which an ordering is important.

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EXAMPLE 6.8

(Lining Up). How many ways are there to put 8 children in a line to get ice cream?

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EXAMPLE 6.8 (Solution)

There are 8 children from which to pick the child who is first.

Then there are 7 children left from which to pick the second child.

Then there are 6 left, and so on.Therefore, there are

8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 = 8! = 40320ways to line up the children.

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THEOREM 6.2

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DEFINITION 6.1

A permutation of k objects from a set of size n is an ordered list of k of the n objects.

The number of permutations of k objects from n is denoted P(n, k).

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THEOREM 6.3

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Combinations

DEFINITION 6.2

A combination of k elements from a set of size n is a subset of size k.

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THEOREM 6.4

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EXAMPLE 6.18

How many license plates consisting of 6 digits (0 to 9) with exactly 2 of the same digit are possible?

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There are 10 digits from which to choose the repeated one.

There are ways to choose the two positions to contain the repeated digit.

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2

6


Then there are P(9, 4) ways to fill in the 4 remaining distinct digits.

By the Multiplication Principle, there are

6-digit license plates with exactly 2 digits the same.

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6.3 Addition and Subtraction

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6.3 Addition and Subtraction

EXAMPLE 6.20

How many possible license plates consisting of 6 digits (0 to 9) have either all digits distinct or all digits the same?

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There are P(10, 6) plates with all digits distinct and 10 with all digits the same.

Certainly no one plate can have both of these properties. Hence, the total number of license plates under consideration is

P(10, 6) + 10 = 151210.

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THEOREM 6.6

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EXAMPLE 6.26

A byte is a binary number consisting of 8 digits. How many bytes have at least 2 zeros?

There are 28 binary sequences of length 8.

Of them, 1 has no zeros and 8 have one zero.

Since the rest have at least 2 zeros, there are

28 − (1 + 8) = 247

sequences with at least 2 zeros.

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THEOREM 6.7

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EXAMPLE 6.29

A standard (6-sided) die is rolled a sequence of 5 times. In how many ways can the sequence of numbers resulting be all even or all multiples of 3?

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Since there are 3 even values (2, 4, and 6), there are 35 ways to get all even numbers.

Since there are 2 multiples of 3 (3 and 6), there are 25 ways to get all multiples of 3.

Only the value 6 is both even and a multiple of 3, so there is 1 way to do both .

Therefore, the desired number of ways is

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6.4 Probability

EXAMPLE 6.31

Consider the task of tossing two dice and recording the numbers showing. Compute the probability that the sum is 8.

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EXAMPLE 6.31 (Cont.)

The set of possible results of our task is thus

S = { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6,

6) }.

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From the possible results listed in S, we are interested in the subset

E = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}.

The probability of E is given by

.

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DEFINITION 6.3

Let S be a finite sample space. The outcomes

in S are said to be equally likely if ∀ x, y ∈S, P(x) = P(y).

That is, ∀x ∈ S, P(x) =

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.||

1

s

DEFINITION 6.4

If the outcomes in a finite sample space S are all equally likely, then the probability of an event E is given by

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||

||)(

S

EEP

THEOREM 6.9

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EXAMPLE 6.34

The license plates issued by a certain state consist of 3 letters (A to Z) followed by 4 digits (0 to 9). If all possible plates are equally likely to be chosen, then what is the probability that a randomly chosen plate will have some letter or digit repeated?

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The sample space S is the set of all possible plates consisting of 3 letters followed by 4 digits. Hence,

|S| = 263 · 104 = 175760000.

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Let E be the subset of plates in which some letter or digit is repeated.

Clearly, |Ec| = P(26, 3) · P(10, 4) = 78624000.

The desired probability is

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THEOREM 6.10

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EXAMPLE 6.36

(Home Security). A home security code consists of 4 digits (0 to 9). If all such codes are equally likely to be chosen, then what is the probability that a randomly chosen code will contain exactly 1 three or exactly 2 sixes?

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Let S be the sample space of all possible 4-digit codes. Let E be the subset of codes containing exactly 1 three, and let F be the subset of codes containing exactly 2 sixes.

So E ∪ F consists of all codes containing exactly 1 three or exactly 2 sixes.

We seek P(E ∪ F).

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Since

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Theorem 6.10 gives that

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Conditional Probability

DEFINITION 6.5

Let E and F be events in a sample space S with P(F) > 0. The conditional probability of E given F , denoted P(E | F ), is given by

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.)(

)()|(

FP

FEPFEP

EXAMPLE 6.37

At a company picnic, the children played a soccer game, after which one player’s name was randomly drawn to win a prize.

The winning team in the soccer game consists of 7 girls and 4 boys, and the losing team consists of 5 girls and 6 boys.

Given that the prize winner is a boy, what is the probability that he also comes from the winning soccer team?

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Let S be the set of 22 children that played soccer, let E be the event that the prize winner is from the winning team, and let F be the event that the prize winner is a boy.

The probability that we seek is P(E|F).

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Note that E ∩ F is the event that the prize winner is a boy from the winning team.

Clearly,

We conclude that

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DEFINITION 6.6

Two events E and F in a sample space S are said to be independent if

P(E ∩ F ) = P(E) · P(F ).

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EXAMPLE 6.38

Consider the experiment of tossing two dice in sequence. Let E be the event that the first die shows a value of at most 3, let F1 be the event that the values on the two dice are the same, and let F2 be the event that the sum of the values on the two dice is 5.

Note that

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(a) Are E and F1 independent?

Solution.

and .

It follows that

So yes, E and F1 are independent.

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(b) Are E and F2 independent?

Solution.

and .

It follows that

So no, E and F2 are not independent.

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THEOREM 6.11

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THEOREM 6.11 -- Proof

E = E ∩ S = E ∩ (F1 ·· · ∪ ∪ Fn) = (E ∩ F1) ·· ∪· (∪ E ∩ Fn) is a disjoint union.

It follows that

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COROLLARY 6.12

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EXAMPLE 6.39

In a recent election, the majority of female voters preferred a different candidate from the one preferred by the majority of the male voters.

Exit polls showed that 75% of female voters chose candidate A, whereas 55% of male voters chose candidate B.

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Assume that each voter chose either candidate A or candidate B and that an equal number of men and women voted.

(a) Which candidate won the election? Explain.

(b) If a randomly chosen voter is known to have voted for candidate A, then what is the probability that the voter is a female?


To be simple, let A be the event that a voter chooses candidate A, let B be the event that a voter chooses candidate B, let F be the event that a voter is female, and let M be the event that a voter is male.

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Based on the exit polls and the assumption that each voter chose A or B, we know that

P(A | F) = .75, P(B | F) = .25,

P(A | M) = .45, P(B | M) = .55.

We are also assuming that

P(F) = P(M) = .5.

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(a) To determine who won, we apply Theorem 6.11. Observe that

P(A) = P(A | F)P(F) + P(A | M)P(M)

= (.75)(.5) + (.45)(.5) = .6.

Since candidate A received 60% of the votes, candidate A won the election.

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(b) We seek P(F | A), the probability that a voter is female given that she voted for candidate A.

For this, Bayes’ Formula gives that

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6.5 Applications of Combinations (Choices with Repetition)

EXAMPLE 6.42 (Barbecue Orders).A barbecue is attended by 7 people.Each person has the choice of a hamburger, a piece of

barbecued chicken, or a hot dog (but only one in each case) for the first food item.

If the cook will barbecue all 7 items at the same time and does not care who ordered what, then how many different barbecue orders are possible for the cook? We assume that there are ample supplies of each food type.

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This is considered a problem involving choices with repetition, since the cook may receive multiple requests for any of the 3 items:

hamburger B, barbecued chicken C, or hot dog D.

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A total order would consist of a list of length 7 of B’s, C’s, and D’s, but with no specified number of any particular choice. For example, one order might be CBBDBDB. To help motivate our counting technique, such an order could be recorded in the form of a table.

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Imagine that the cook simply places a in the appropriate column of an order sheet as each order is taken. At the end, the order could more efficiently be recorded as a sequence of length 9 consisting of 7 ’s and 2 |’s.

| |

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Here the headers B, C, D are understood. The point is that this is a binary sequence of length 9 containing 7 ’s (and 2 |’s). Note that the number of |’s is one fewer than the number of item choices. We conclude that the number of possible barbecue orders is

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THEOREM 6.13

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6.6 Correcting for Overcounting

EXAMPLE 6.44 (Seating Arrangements).

How many ways are there to seat 5 girls at a circular table if the particular seat taken by each girl does not matter and what matters to each girl is

(a) who is sitting to her left and who is sitting to her right?

(b) who is sitting next to her (which side does not matter)?

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Temporarily call one seat the head of the table.If we keep track of which girl is seated at the head, then

there are 5! ways to seat the girls clockwise around the table.

However, 5! is an overcounting of what we want, since we have carried the extra structure of who is seated at the head of the table.

Given any such seating, if all of the girls stood up and shifted one position clockwise, then the new seating should be considered the same as the original.

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(a) Each girl would still have the same neighbor to her left and the same neighbor to her right. Since there are 5 different rotations of any seating, we need to divide the original 5! count by 5.

Therefore, there are different seatings around the table.

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(b) When which neighbor is to the left of a girl and which is to the right no longer matters, there are fewer than 24 different seatings.

Since only the set of 2 neighbors matters, if we take the mirror reflection of any seating from part (a), then the reflected seating should now be considered the same as the original.

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Reflection preserves neighbors (and switches sides).

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Since each seating from part (a) should be paired with its reflection, we need to divide the answer from part (a) by 2.

Thus, there are different seatings in which only neighbors matter.

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