Ch3 Bijection Principle

Raffles Institution H3 Mathematics 9810 (Combinatorics) Year 6 2012

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__________________________ Chapter 3 The Bijection Principle

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Chapter 3: The Bijection Principle

1 Injective and Surjective Mappings

Let us first recall some concepts on mappings of sets.

A mapping f from A to B, denoted by f : A B→ , is a rule which assigns to each element a in

A a unique element, denoted by f(a) in B. f(a) is also called as the image of a and a is the

preimage of f(a).

1.1 Injective Mappings

We say that a mapping is injective (or one to one) if f(x)≠f(y) in B whenever x≠ y in A.

Example:

a

1 b

3 c

5 d

A B

Qn: Can you modify f so that it is not one to one?

1.2 Surjective mappings We say that a mapping is surjective (or onto) if for any b in B, there exists an a in A such that

f(a) = b, i.e. every element in b has a preimage in A.

Example:

a

1 b

3 c

5 d

A B

a

b

c

d

c does not have a preimage.

Thus f1 is not onto.

a, b, c, d all have a preimage.

Thus f2 is onto.

Each element in A corresponds

to distinct images in B.

Thus f is one to one.

f

f1

f2

A B

Raffles Institution H3 Mathematics 9810 Year 6 2012 __________________________________________

___________________________ Chapter 3 The Bijection Principle

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Note: Let A and B be finite sets.

(a) If there exists an injective mapping from A to B , clearly BA ≤≤≤≤ .

(b) If there exists a surjective mapping from A to B , then BA ≥≥≥≥ .

2 The Bijection Principle

A bijective mapping is both injective and surjective. Thus, we have

By establishing a bijection, we can reduce a difficult or unfamiliar counting problem to

something easy and familiar. More often than not, we will reduce our problem to that of

counting sequences or sets.

Example 1

An ant wants to crawl from P to Q. If it can only crawl along the segments of the grid, how

many shortest routes does it have to get from P to Q?

Q

P

Solution: Instead of counting manually, we could use 0 to represent a horizontal segment of the route,

and 1 to represent a vertical segment. For example,

Q

P

The route above could be represented by 00110001 – a binary sequence.

Similarly, the route Q

P

could be represented by 10001100 – another binary sequence.

The Bijection Principle (BP)

Let A and B be finite sets. If there exists a bijection f : A B→ , then

.A B=



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Let A be the set of all possible shortest P-Q routes and B the set of all 8-digit binary

sequences with three 1’s. The above representation of a shortest route by a binary sequence

where a horizontal segment of the route is represented by a “0” while a vertical segment is

represented by a “1” defines a mapping f from A to B, namely, f : A B→ .

Clearly, different shortest P-Q routes in A correspond to different sequences in B under f, so f

is one to one. For any sequence b in B, one can find a corresponding shortest P-Q route in A,

so f is onto.

Thus, by (BP), |A| = |B| = 8

563

=

.

Example 2 Find the number of divisors of 12600.

Solution: To solve this problem of counting, we have to make use of the Fundamental Theorem of

Algebra.

Every positive integer n ≥ 2 can be factorized as 1 2

1 2 ... kmm m

kn p p p=

for some prime numbers p1, p2, …, pk and for some positive integers m1, m2, …, mk.

Such a factorization is unique if the order of primes is disregarded.

Prime decomposition of 12600 gives 3 2 2 112600 2 3 5 7= × × × .

Thus a number d divides 12600 if and only if it can be expressed as

where p, q, r, s are integers such that 0 3,0 2,0 2,0 1p q r s≤ ≤ ≤ ≤ ≤ ≤ ≤ ≤ .

Let A be the set of divisors of 12600 and

{( , , , ) : 0,1, 2,3; 0,1, 2; 0,1, 2;B p q r s p q r= = = = 0,1}s = .

We define the mapping f : A B→ as f( ) ( , , , ),d p q r s d A= ∈ .

To prove f is a bijection from A to B:

Suppose d1≠d2, then f(d1)= f(d2) which contradicts the Fundamental Theorem of Algebra. So

f is one to one.

2 3 5 7p q r s× × × divides 12600 for 0 3,0 2,0 2,0 1p q r s≤ ≤ ≤ ≤ ≤ ≤ ≤ ≤ .

So ( , , , ) f( )p q r s d= for some 2 3 5 7p q r sd = × × × .

Hence, f is onto.

Since f is a bijection, by (BP) and (MP), 4 3 3 2 72A B= = × × × = .

So far the bijections we have formed are with sequences. Let’s consider the next example,

where the bijection formed is with subsets:

2 3 5 7p q r sd = × × ×



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Example 3 There are five distinct points on a circle. How many chords can be formed by joining two of

these points?

Solution:

Let A be the set of such chords, and B the set of 2-element subsets of {1, 2, 3, 4, 5}. For each

chord α in A, define f( ) { , }p qα = , where p, q are the two points on the circle chosen to form

the chord α. If α and β are two distinct chords, then ( ) ( )f fα β≠ which implies f is injective.

For any 2-element subset {p, q} in B, there is a chord α in A that is formed by joining the two

points p and q. So f is onto.

Hence, f : A B→ is a bijection; by (BP), .A B=

Since 5

2B

=

, 5

2A

=

= 10, which implies a total of 10 chords can be formed.

Extension of Example 3: Consider the 10 chords that can be formed in Example 3. Given that no three chords are

concurrent, find the number of points of intersection of these 10 chords within the circle.

Solution: Once again we make use of (BP).

Let C be the set of all points of intersection between the 10 chords and D the set of 4-element

subsets of {1, 2, 3, 4, 5}. The following figure exhibits a rule between C and D.

For each point β in C, define g (β) = {p, q, r, s }, where p, q, r, s are points on the

circumference such that they are the endpoints of the two chords that intersect to give β. So, g

is a mapping from C to D.

4

1 2

3 5

{1, 2,4,5}

{1,2,3,5}

{1,2,3,4}

{2,3,4,5}

{1,3,4,5}

a

b

c

d

e

�

�

�

�

�

4

2

3 5

a

b

c

d e



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Every pair of chords, with endpoints p, q, r, s, that intersect, will result in a point of

intersection, so g is surjective (onto). Also, two distinct points of intersection will not have

the same pair of chords, so g is injective (one to one).

∴ g: C → D is a bijection.

Once we have done that, we know .C D= Since 5

4D

=

, thus 5

4C

=

.

Example 4

The sets S, T and U are defined as follows:

S = {1, 2, 3, …, 2n, 2n + 1},

T = set of subsets which contain 0, 2, 4, …, 2n elements of S,

U = set of subsets which contain 1, 3, 5, …, (2n + 1) elements of S,

where n is a positive integer.

For example, if S = {1 , 2, 3}, then T = {{ ∅}, {1, 2}, {1, 3}, {2, 3}} and

U = {{1}, {2}, {3}, {1, 2, 3}}.

By using the bijection principle, show that | |T = | |U .

Hence write down the number of subsets which contain an even number of elements for a set

with 9 elements. [4]

Solution:

We define a mapping f: T → U such that

Example 5

Given that X = {1, 2, …, n} where n∈ +� , show that the number of combinations of X

containing r elements with no two elements being consecutive integers is 1n r

r

− +

, where

r∈ +� and 0 1r n r≤ ≤ − + .

Solution:

Let A be the set of r-combinations of X with no two elements being consecutive integers and

B be the set of r-combinations of Y, where Y = {1, 2, …, n − r+1 }.

For any a∈A, a = {a1, a2, …, ar} where we may assume1 2 ... ra a a< < < . We define a mapping

f: A →B such that f(a) =



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3 Exercise

1. Power sets

The power set of a set, denoted by P(S), is the set of all subsets of S, including the

empty set φ and S itself. For example, if {1, 2, , }, 1 3nN n n= ≤ ≤… , then

1( ) { ,{1}},P N φ=

2( ) { ,{1},{2},{1, 2}},P N φ=

3( ) { ,{1},{2},{3},{1, 2},{1,3},{2,3},{1,2,3}}.P N φ=

We see that 1( ) 2P N = , 2

( ) 4P N = , 3( ) 8P N = . Do you think there is a definite

pattern? Can you form a bijection with binary sequences to find ( )n

P N ?

2. Ordered decomposition of positive integers

The number 4 can be expressed as a sum of one or more positive integers, taking

order into account, in the following eight ways:

4 4

1 3

3 1

2 2

1 1 2

1 2 1

2 1 1

1 1 1 1

=

= +

= +

= +

= + +

= + +

= + +

= + + +

Find the number of ways the positive integer n can be expressed in.

3. Find the number of parallelograms which are contained in the configuration of the

equilateral triangle below and which have no sides parallel to BC. (Hint: Adjoin a

new row at the base of the triangle.)

4. In a sequence of coin tosses, HT is used to denote the result when a head is followed

by a tail. In the sequence THTTHHTHHTTTTTH for 15 coin tosses, we observe that

there are exactly two HH, three HT, four TH and five TT subsequences. How many

sequences of 15 coin tosses will contain exactly two HH, three HT, four TH and five

TT subsequences?

B C

A



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Ch3 Bijection Principle

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