R. Johnsonbaugh Discrete Mathematics 5 th edition, 2001
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Transcript of R. Johnsonbaugh Discrete Mathematics 5 th edition, 2001
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R. Johnsonbaugh
Discrete Mathematics 5th edition, 2001
Chapter 4Counting methods
and the pigeonhole principle
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4.1 Basic principlesMultiplication principle
If an activity can be performed in k successive steps, Step 1 can be done in n1 ways
Step 2 can be done in n2 ways … Step k can be done in nk ways
Then: the number of different ways that the
activity can be performed is the product
n1n2…nk
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Addition principleLet X1, X2,…, Xk be a collection of k pairwise
disjoint sets, each of which has nj elements, 1 < j < k, then the union of those sets
k
X = Xj
j =1
has n1 + n2 + … + nk elements
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4.2 Permutations and combinations
A permutation of n distinct elements x1, x2,…, xn is an ordering of the n elements. There are n! permutations of n elements.
Example: there are 3! = 6 permutations of three elements a, b, c:
abc bac cab
acb bca cba
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r-permutationsAn r-permutation of n distinct elements is an
ordering of an r-element subset of the n elements x1, x2,…, xn
Theorem 4.2.10:
For r < n the number of r-permutations of a set with n distinct objects is
P(n,r) = n(n-1)(n-2)…(n-r+1)
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Combinations
Let X = {x1, x2,…, xn} be a set containing n
distinct elements An r-combination of X is an unordered
selection of r elements of X, for r < n The number of r-combinations of X is the
binomial coefficient
C(n,r) = n! / r!(n-r)! = P(n,r)/ r!
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Catalan numbers
Eugene-Charles Catalan (1814-1894)
Catalan numbers are defined by the formula
Cn = C(2n,n) / (n+1)
for n = 0, 1, 2,…
The first few terms are:
n 0 1 2 3 4 5 6 7 8 9 10
Cn 1 1 2 5 14 42 132 429 1430 4862 16796
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4.3 Algorithms for generating permutations and combinations
Lexicographic order: Given two strings = s1s2…sp and = t1t2…tq
Define < if p < q and si = ti for all i = 1, 2,…, p
Or for some i, si ti and for the smallest i, si < ti
Example: if = 1324, = 1332, = 132,
then < and < .
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4.4 Introduction to discrete probability
An experiment is a process that yields an outcome
An event is an outcome or a set of outcomes from an experiment
The sample space is the event of all possible outcomes
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Probability
Probability of an event is the number of outcomes in the event divided by the number of outcomes in the sample space.
If S is a finite sample space and E is an event (E is a subset of S) then the probability of E is
P(E) = |E| / |S|
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4.5 Discrete probability theory
When all outcomes are equally likely and there are n possible outcomes, each one has a probability 1/n.
BUT this is not always the case. When all probabilities are not equal, then some probability (possibly different numbers) must be assigned to each outcome.
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Probability function A probability function P is a function from
the set of all outcomes (sample space S) to the interval [0, 1], in symbols
P : S [0, 1]
The probability of an event E S is the sum of the probabilities of every outcome in E
P(E) = P(x)
x E
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Probability of an event Given E S, we have
0 < P(E) < P(S) = 1
If S = {x1, x2,…, xn} is a sample space, then
n
P(S) = P(xi) = 1
i =1
If Ec is the complement of E in S, then
P(E) + P(Ec) = 1
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Events in a sample space
Given any two events E1 and E2 in a sample space S. Then
P(E1 E2) = P(E1) + P(E2) – P(E1E2)
We also have P() = 0
Events E1 and E2 are mutually exclusive if and
only if E1E2 = . In this case
P(E1E2) = P(E1) + P(E2)
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Conditional probability
Conditional probability is the probability of an event E, given that another event F has occurred, is called. In symbols P(E|F).
If P(F) > 0 then
P(E|F) = P(EF) / P(F) Two events E and F are independent if
P(EF) = P(E)P(F)
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Pattern recognition Pattern recognition places items into classes,
based on various features of the items. Given a set of features F we can calculate the
probability of a class C, given F: P(C|F) Place the item into the most probable class, i.e.
the one C for which P(C|F) is the highest. Example: Wine can be classified as Table wine (T), Premium (R)
or Swill (S). Let F {acidity, body, color, price} Suppose a wine has feature F, and P(T|F) = 0.5, P(R|F) = 0.2
and P(S|F) = 0.3. Since P(T|F) is the highest number, this wine will be classified as table wine.
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Bayes’ Theorem Given pairwise disjoint classes C1, C2,…,
Cn and a feature set F, then
P(Cj|F) = A / B, where
A = P(F|Cj)P(Cj)
n
and B = P(F|Ci)P(Ci)
i = 1
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Generalized permutations and combinations
Theorem 4.6.2: Suppose that a sequence of n items has nj identical objects of type j,
for 1< j < k. Then the number of orderings of S is
____n!____
n1!n2!...nk!
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4.7 Binomial coefficients and combinatorial identities
Theorem 4.7.1: Binomial theorem. For any real numbers a, b, and a nonnegative integer n: (a+b)n = C(n,0)anb0 + C(n,1)an-1b1 + … + C(n,n-1)a1bn-1 + C(n,n)a0bn
Theorem 4.7.6: For 1 < k < n, C(n+1,k) = C(n,k) + C(n,k-1)
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Pascal’s Triangle 1 1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
…
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4.8 The pigeonhole principle First form: If k < n and n pigeons fly into k
pigeonholes, some pigeonhole contains at least two pigeons.
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Second form of the pigeonhole principle
If X and Y are finite sets with |X| > |Y| and f : X Y is a function, then f(x1) = f(x2) for some x1, x2 X, x1 x2.
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Third form of the pigeonhole principle
If X and Y are finite sets with |X| = n, |Y| = m and k = n/m, then there are at least k values a1, a2,…, ak X such that f(a1) = f(a2) = … f(ak).
Example: n = 5, m = 3k = n/m = 5/3 = 2.