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Applications of Set Theory in Counting Techniques: Inclusion-Exclusion RulePowerPoint Presentation

Applications of Set Theory in Counting Techniques: Inclusion-Exclusion Rule

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Applications of Set Theory in Counting Techniques: Inclusion-Exclusion Rule

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Applications of Set Theory in Counting Techniques: Inclusion-Exclusion Rule

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Applications of Set Theory in Counting Techniques: Inclusion-Exclusion Rule

C

A

A

B

B

- If A, B and C are finite sets then
n(A B) = n(A) + n(B) – n(A B)

n(A B C) = n(A) + n(B) + n(C)

- n(A B) – n(A C) – n(B C)

+ n(A B C)

- Question: How many integers from 1 through 100
are multiples of 3 or multiples of 7 ?

- Solution: Let A=the set of integers from 1 through 100 which are multiples of 3;
B = the set of integers from 1 through 100

which are multiples of 7.

Then we want to find n(A B).

First note that A B is the set of integers

from 1 through 100 which are multiples of 21 .

n(A B) = n(A) + n(B) - n(A B) (by incl./excl. rule)

= 33 + 14 – 4 = 43 (by counting the elements

of the three lists)

- 3 headache drugs – A,B, and C – were tested on 40 subjects. The results of tests:
23 reported relief from drug A;

18 reported relief from drug B;

31 reported relief from drug C;

11 reported relief from both drugs A and B;

19 reported relief from both drugs A and C;

14 reported relief from both drugs B and C;

37 reported relief from at least one of the drugs.

Questions:

1) How many people got relief from none of the drugs?

2) How many people got relief from all 3 drugs?

3) How many people got relief from A only?

C

A

B

S

We are given: n(A)=23, n(B)=18, n(C)=31,

n(A B)=11, n(A C)=19, n(B C)=14 ,

n(S)=40, n(A B C)=37

Q1) How many people got relief from none of the drugs?

By difference rule,

n((A B C)c ) = n(S) – n(A B C) = 40 - 37 = 3

Q2)How many people got relief from all 3 drugs?

By inclusion/exclusion rule:

n(A B C) = n(A B C)

- n(A) - n(B) - n(C)

+ n(A B) + n(A C) + n(B C)

= 37 – 23 – 18 – 31 + 11 + 19 + 14 = 9

Q3)How many people got relief from A only?

n(A – (B C)) (byinclusion/exclusion rule)

= n(A) – n(A B) - n(A C) + n(A B C)

= 23 – 11 – 19 + 9 = 2