Basic Combinatorics Inclusion-Exclusion Principle Clarification

Click For Summary

Homework Help Overview

The discussion revolves around a combinatorial problem involving the Inclusion-Exclusion Principle to determine the count of integers between 1 and 2009 that are not divisible by certain numbers (3, 2, and 10). Participants explore two parts of the problem, focusing on different interpretations of the conditions set forth.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants present their calculations and reasoning based on the Inclusion-Exclusion Principle, questioning the relationships between their results for parts (a) and (b). There is a discussion about whether the counts for integers not divisible by all three numbers should be greater than those not divisible by any of them.

Discussion Status

Some participants have provided alternative interpretations of the calculations, leading to a re-evaluation of the answers. There is an ongoing examination of the logic behind the application of the Inclusion-Exclusion Principle, with no clear consensus reached on the correct answers yet.

Contextual Notes

Participants are navigating the complexities of the Inclusion-Exclusion Principle and its application to the specific problem constraints. There is a noted confusion regarding the definitions of the sets involved and how they relate to the overall counts.

pupeye11
Messages
99
Reaction score
0

Homework Statement



How many integers between 1 and 2009, inclusive are (a) not divisible by 3,2, and 10 (b) not divisible by 3,2, or 10?

Homework Equations



The number of objects of the set S that have none of the properties is given by the alternating expression:

[tex] \mid S \mid -\sum \mid A_{i} \mid + \sum \mid A_{i} \cap A_{j} \mid - ... [/tex]

The number of objects of S which have at least one of the properties is given by

[tex] \sum \mid A_{i} \mid - \sum \mid A_{i} \cap A_{j} \mid + ... [/tex]


The Attempt at a Solution



Well i=1,2,3 where i=1 is the property that its divisible by 3 and i=2 for the property that its divisible by 2 and i=3 for the property its divisible by 10

[tex] \mid S \mid = 2009<br /> \mid A_{1} \mid = 669<br /> \mid A_{2} \mid = 1004<br /> \mid A_{3} \mid = 200<br /> \mid A_{1} \cap A_{2} \mid = 334<br /> \mid A_{1} \cap A_{3} \mid = 66<br /> \mid A_{2} \cap A_{3} \mid = 100<br /> \mid A_{1} \cap A_{2} \cap A_{3} \mid = 66[/tex]

So for the part a I thought it would be the first equation because I am looking for the number of objects that don't have all these properties at the same time so my answer came to 570.

For part b I thought it'd be the second equation which comes to 1439.

But shouldn't the number of integers that are not divisible by 3,2, and 10 be larger than the number of integers that are not divisible by 3,2, or 10?
 
Physics news on Phys.org
Ok, so pretty much I just had my two answers backwards. 570 is the answer to (b) by the first equation because I found an example and I did it the exact same way. Which makes part (a) equal to 1439 by the second equation. Correct?
 
(a) [tex]\sim{A_1} \cap \sim{A_2} \cap \sim{A_3}=\sim (A_1 \cup A_2 \cup A_3)=2009-1439=570[/tex]

(b)[tex]\sim{A_1} \cup \sim{A_2} \cup \sim{A_3}=\sim (A_1 \cap A_2 \cap A_3)=2009-66=1943[/tex]
 
Don't you have those backwards though.

(a) not divisible by 3,2, and 10 which is just
[tex] \sim (A_1 \cap A_2 \cap A_3)=2009-66=1943[/tex]

so obviously the 2009 - 66 would make sense here and

(b) not divisible by 3,2, or 10
would be the other method.
 
And what made you think that I did those backwards?

Just follow what the task is asking for.

Regards.
 

Similar threads

  • · Replies 13 ·
Replies
13
Views
4K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 31 ·
2
Replies
31
Views
3K
Replies
9
Views
2K
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K