Jong C. Park

Computer Science Division, KAIST

Discrete Mathematics, 2008 Computer Science Division, KAIST2

Today’s TopicsBasic Principles

Permutations and Combinations

Algorithms for Generating Permutations

Generalized Permutations and Combinations

Binomial Coefficients and Combinatorial Identities

The Pigeonhole Principle

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• Definition

– Given a set X = {x1, ..., xn} containing n

(distinct) elements,

(a) An r-combination of X is an unordered selection

of r-elements of X (i.e., an r-element subset of X).

(b) The number of r-combinations of a set of n

distinct elements is denoted C(n,r) or (n r).

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• Theorem– The number of r-combinations of a set of n

distinct objects is C(n,r) = P(n,r)/r! = n(n – 1)···(n– r + 1)/r! = n!/((n - r)!·r!), r n.

• Examples– In how many ways can we select a committee of

three from a group of 10 distinct persons?• C(10,3)

– In how many ways can we select a committee of two women and three men from a group of five distinct women and six distinct men?

• C(5,2)·C(6,3) = 10·20 = 200

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• Examples

– How many eight-bit strings contain exactly four 1’s?• C(8,4) = 70

– An ordinary deck of 52 cards consists of four suits (clubs, diamonds, hearts, spades) of 13 denominations each (ace, 2-10, jack, queen, king).

(a) How many (unordered) five-card poker hands, selected from an ordinary 52-card deck, are there?

C(52,5) = 2,598,960

(b) How many poker hands contain cards all of the same suit?

4·C(13,5) = 5148

(c) How many poker hands contain three cards of one denomination and two cards of a second denomination?

13·C(4,3)·12·C(4,2) = 3744

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• Examples– How many routes are there from the lower-left

corner of an n n square grid to the upper-right corner if we are restricted to traveling only to the right or upward?

• C(2n,n)

– How many routes are there from the lower-left corner of an n n square grid to the upper-right corner if we are restricted to traveling only to the right or upward and if we are allowed to touch but not go above a diagonal line form the lower-left corner or the upper-right corner?

• C(2n,n) – C(2n,n-1) = C(2n,n)/(n + 1)

• Catalan numbers: C0 = 1, C1 = 1, C2 = 2, C3 = 5, …

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• Note

– The rock group “Unhinged Universe” has recorded nvideos whose running times are t1, t2, ..., tn seconds.

– A tape is to be released that can hold C seconds. Since this is the first tape by the Unhinged Universe, the group wants to include as much material as possible.

– The problem is to choose a subset {i1, ..., ik} of {1, 2, ..., n} such that the sum k

j=1tijdoes not exceed C

and is as large as possible.

– For this purpose, we can examine all subsets of {1, 2, …, n} and choose a subset so that the sum does not exceed C and is as large as possible.

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• Definition– Let = s1s2···sp and = t1t2···tq be strings over {1,

2, ···, n}.– We say that is lexicographically less than

and write < if either (a) p < q and si = ti for i = 1, ..., p, or (b) for some i, si ti, and for the smallest such i, we have si < ti.

• Examples– Determine the lexicographic relation between

and as defined below.• = 132 and = 1324. • = 13246 and = 1342.• = 1324 and = 1342.• = 13542 and = 21354.

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• Examples– Consider the (lexicographic) order in which the

5-combinations of {1, 2, 3, 4, 5, 6, 7} will be listed. What are the first string, the next string, and the last string?

• 12345, 12346, and 34567

– Find the string that follows 13467 when we list the 5-combinations of X = {1, 2, 3, 4, 5, 6, 7}.

• 13567

– Find the string that follows 2367 when we list the 4-combinations of X = {1, 2, 3, 4, 5, 6, 7}.

• 2456

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11

12

13

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• Example– How many strings can be formed using the

following letters?M I S S I S S I P P I

• Solution– Consider the problem of filling 11 blanks with the letters

given:

_ _ _ _ _ _ _ _ _ _ _.

– First, there are C(11,2) ways to choose positions for the two P’s.

– Then, there are C(9,4) ways to choose the remaining positions for the four S’s.

– Finally, there are C(5,4) ways to choose positions for the four I’s.

– C(11,2)·C(9,4)·C(5,4) = (11!/(2!·9!))·(9!/(4!·5!))·(5!/(4!·1!)) = 11!/(2!·4!·4!·1!) = 34,650.

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• Theorem

– Suppose that a sequence S of n items has n1

identical objects of type 1, n2 identical objects of

type 2, ..., and nt identical objects of type t. Then

the number of orderings of S is n!/(n1!·n2!···nt!).

• Example

– In how many ways can eight distinct books be

divided among three students if Bill gets four

books and Shizuo and Marian each get two

books?

• 8!/(4!·2!·2!) = 420

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• Example

– Consider three books: a computer science

book, a physics book, and a history book.

Suppose that the library has at least six

copies of each of these books. In how many

ways can we select six books?

• C(8,2) = 28

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• Theorem– If X is a set containing t elements, the number of

unordered, k-element selections from X, repetitions allowed, is C(k + t – 1, t – 1) = C(k + t– 1, k).

• Example– Suppose that there are piles of red, blue, and

green balls and that each pile contains at least eight balls.

(a) In how many ways can we select eight balls?– C(8+3-1,3-1) = C(10,2) = 45

(b) In how many ways can we select eight balls if we must have at least one ball of each color?

– C(5+3-1,3-1) = C(7,2) = 21

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• Examples

– In how many ways can 12 identical mathematics books be distributed among the students Anna, Beth, Candy, and Dan?

• C(12+4-1,4-1) = C(15,3) = 455

– How many solutions in nonnegative integers are there to the equation x1 + x2 + x3 + x4 = 29?

• C(29+4-1,4-1) = C(32,3) = 4960

– How many solutions in integers are there in the equation above satisfying x1 > 0, x2 > 1, x3 > 2, x4 0?

• C(23+4-1,4-1) = C(26,3) = 2600

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• Note

– (a+b)3 = (a+b)(a+b)(a+b)

= aaa + aab + aba + abb + baa + bab + bba + bbb

= a3 + a2b + a2b + ab2 + a2b + ab2 + ab2 + b3

= a3 + 3a2b + 3ab2 + b3

– (a+b)n = C(n,0)anb0

+ C(n,1)an-1b1

+ C(n,2)an-2b2

+ ···

+ C(n,n-1)a1bn-1

+ C(n,n)a0bn

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• Theorem (Binomial Theorem)– If a and b are real numbers and n is a positive integer,

then (a + b)n = nk=0C(n,k)an-kbk.

– The numbers C(n,r) are known as binomial coefficients.

• Examples– Obtain (a + b)3 using the Binomial Theorem.

– Expand (3x – 2y)4 using the Binomial Theorem.

– Find the coefficient of a5b4 in the expansion of (a + b)9.

• C(9,4) = 126

– Find the coefficient of x2y3z4 in the expansion of (x + y + z)9.

• C(9,2)·C(7,3) = 9!/(2!·3!·4!) = 1260

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• Note

– Pascal’s triangle

• The border consists of 1’s, and any interior value

is the sum of the two numbers above it.

– Combinatorial identity

• An identity that results from some counting

process

– Combinatorial argument

• The argument that leads to the formulation of a

combinatorial identity

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• Theorem (Pascal’s Triangle)– C(n + 1,k) = C(n,k – 1) + C(n,k) for 1 k n.

– Proof.• Let X be a set with n elements. Choose a X. Then

C(n + 1,k) is the number of k-element subsets of Y = X {a}. Now the k-element subsets of Y can be divided into two disjoint classes.

1. Subsets of Y not containing a.

2. Subsets of Y containing a.

• The subsets of class 1 are just k-element subsets of Xand there are C(n,k) of these. Each subset of class 2 consists of a (k – 1)-element subset of X together with a and there are C(n,k – 1) of these.

• Therefore, C(n + 1,k) = C(n,k – 1) + C(n,k).

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• Examples

– Use the Binomial Theorem to derive the

equation nk=0C(n,k) = 2n.

– Show that ni=kC(i,k) = C(n + 1,k + 1).

– Use the equation above to find the sum 1 + 2

+ ··· n.

• 1 + 2 + ··· n

= C(1,1) + C(2,1) + ... + C(n,1)

= C(n + 1,2)

= (n + 1)n/2

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• Pigeonhole Principle (First Form)– If n pigeons fly into k pigeonholes and k < n,

some pigeonhole contains at least two pigeons.

• Example– Ten persons have first names Alice, Bernard,

and Charles and last names Lee, McDuff, and Ng. Show that at least two persons have the same first and last names.

• Proof.– There are nine possible names for the 10 persons. If we

think of the persons as pigeons and the names as pigeonholes, we can consider the assignment of names to people to be that of assigning pigeonholes to the pigeons. By the Pigeonhole Principle, some name (pigeonhole) is assigned to at least two persons (pigeons).

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• Pigeonhole Principle (Second Form)

– If f is a function from a finite set X to a finite set Yand |X| > |Y|, then f(x1) = f(x2) for some x1, x2 X, x1 x2.

• Examples

– If 20 processors are interconnected, show that at least two processors are directly connected to the same number of processors.

– Show that if we select 151 distinct computer science courses numbered between 1 and 300 inclusive, at least two are consecutively numbered.

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• Pigeonhole Principle (Third Form)– Let f be a function from a finite set X into a finite

set Y. Suppose that |X| = n and |Y| = m. Let k = n/m. Then there are at least k values a1, ..., ak X such that f(a1) = f(a2) = ··· = f(ak).

• Example– A useful feature of black-and-white pictures is

the average brightness of the picture. Let us say that two pictures are similar if their average brightness differs by no more than some fixed value. Show that among six pictures, there are either three that are mutually similar or three that are mutually dissimilar.

• Combinations

• Algorithms for

Generating

Permutations

• Generalized

Permutations and

Combinations

Discrete Mathematics, 2008 Computer Science Division, KAIST27

• Binomial Coefficients

and Combinatorial

Identities

• The Pigeonhole

Principle