SETS & Permutation and Combination

 

Principle of Inclusion and Exclusion

 

Definition: The number of elements in a set A is denoted by .  Hence if set A contains n elements, we write .

 

 

Definition:  We say that 2 sets, A and B, are mutually exclusive if .

 

Hence, given 2 mutually exclusive sets, A and B, we have

 

 

 

 

 

 

 

 

 

 


In general, if A and B are finite sets.  Then

 

 

 

 

 

 

 

 

 


Let A, B and C be finite sets.  Then

 

 

 

 

 

 

 

 

 


Let A, B, C and D be finite sets.  Then

Application 1.         

 

Find the number of integers in  which are divisible by 2,3 or 5.

 

Let A be the subset of S which contains elements that are divisible by 2.

Let B be the subset of S which contains elements that are divisible by 3.

Let C be the subset of S which contains elements that are divisible by 5.

 

 

 

* is the subset of S which contains elements that are divisible by 6

* is the subset of S which contains elements that are divisible by 15

* is the subset of S which contains elements that are divisible by 10

 

 is the subset of S which contains elements that are divisible by 30

 

Hence .

 

 

 

Exercise 1

 

A.        Let A, B and C be finite sets.  Prove that

(i)         ;

(ii)       

where A' and B' are the complements of A and B respectively.

 

B         Apply the result of A(ii) to find the number of integers from 1 to 250, inclusive, which are not divisible by 2 nor by 7 but are divisible by 5.

ANSWER 21

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Application 2

 

Let S be the set of 14 students and be the properties that a students attends the course Philosophy, Economic, Chinese respectively.

 

Student

1

2

3

4

5

6

7

8

9

10

11

12

13

14

4

 

 

4

 

 

 

4

4

 

 

4

 

 

4

4

4

4

 

4

 

4

 

4

4

 

4

 

4

 

4

 

 

 

 

4

4

 

4

 

 

4

 

Let w() denote the number of elements in S satisfying , i =1,2,3

w() means the number of elements in S satisfying both , i,j=1,2,3

 

w() = 5      w() = 9         w() = 6     

w()=3   w()= 3      w()= 4   

w()= 2  

 

The number of students who did not attend any course =

 - { w() +  w() + w() }

     + {w() + w() + w() }

- {w()}

= 14 - 20+ 10-2 = 2

 

The number of students who attended exactly one course =

{ w() +  w() + w() }

-{w() + w() + w()}

+ {w() }

= 20 - 2(10) +3(2) = 6

 

The number of students who attended exactly two courses =

{w() + w() +  w()}

-  {w() }

= 10 - 3(2) =4

The number of students who attended exactly three courses=

 

 

 

 

 

 

 

Exercise 2

 

Find the number of integers from 1 to 100 inclusive, which are divisible by

(a)   non of 2, 3, 5?

(b)   exactly one of 2,3, 5?

(c)   exactly two of 2,3,5?

(d)   all of 2,3,5?

ANSWER 26,48,23,3

 

Let S ={ 1,2,3,….100} andbe the properties that the number is divisible by 2, 3 and 5 respectively.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Application 3

 

Find the number of arrangement using all letters of the word MEGAPHONES which contain no pattern of MA, POH and SHEE.

 

Let S be the set of all arrangement of the given letters.  Let be the properties that the arrangement contains the pattern MA, POH, SHEE respectively.

 

Let w() be the number of elements in S satisfying , i=1,2,3.  That is w() means the number of arrangement which contains the pattern MA.

 

w() means the number of elements in S satisfying , i,j=1,2,3.  That is w() means the number of arrangement which contains both MA and POH.

 

== 1 814 400

w()+ w() + w() ==206 640

w()+w()+w()=+0= 3240

w()= 0

 

Hence the number arrangement using all letters of the word MEGAPHONES which contain no pattern of MA, POH and SHEE =

 - { w()+ w() + w() } + {w()+w()+w() } - w()

=161 100

 

 

Exercise 3

 

Find the number of arrangements using all the letters of the word COMBINATIONS which do not contain any of the pattern BIN, SIN and TOMB.

ANSWER 56185800

 

 

 

 

 

 

 

 

 

 

 

 

Application 4

 

Find the number of integer solutions for the equation  with .

 

(Remark: The answer to this question is trivial intuitively since 5+6+7=18 and , we can make use of set theory to outline a proof which is applicable for  with  where .)

 

 

Let S be the set of all integer solutions for with .

 

Let  be the properties that respectively. 

 

Let w() be the number of elements in S satisfying .

 

It can be shown that = =

 

w() + w()  + w() = =

 

w()+w() + w()==

 

 w() = =

 

The number of integer solutions for the equation  with  is

 - { w() + w()  + w() } + { w()+w() + w()} - w() = - { } +{  } -  = 0

 

 

Exercise 4

 

Find the number of integer solutions for the equation  with .

ANSWER 106

 

Prepared By Ms Khong Beng Choo

Adapted from MQ3207 Combinatorial Analysis,2001