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prove something must happen with pigeonhole principle [duplicate]

If any $15$ cars have $3$ with the same manufacturer, prove that any $100$ cars have $15$ with the same manufacturer.Pigeonhole Principle used for Finding NumbersPigeonhole principle to prove divisionPigeonhole principle in practice?Pigeonhole Principle and SetsWhether it is the pigeonhole principle?Geometry pigeonhole principle problem.pigeonhole principle clarificationCounting problem, probably related with pigeonhole principleDealing with Pigeonhole Principle ProblemsIf any $15$ cars have $3$ with the same manufacturer, prove that any $100$ cars have $15$ with the same manufacturer.

-1

$begingroup$

This question already has an answer here:

• 
  If any $15$ cars have $3$ with the same manufacturer, prove that any $100$ cars have $15$ with the same manufacturer.
  
  3 answers
  
  

I'm given the following problem:

In every group of 15 cars , there are 3 that were manufactured in the same country.
prove that in a group of 100 cars there are 15 cars that were manufactured in the same country.

does anyone have an idea of how to implement pigeonhole principle here? I'm kind of stuck because I can't assign cars to countries since I don't know how many countries are there..
Any help is appreciated . Thanks.

combinatorics pigeonhole-principle

share|cite|improve this question

edited Mar 18 at 21:02

Maria Mazur

49.4k1361124

asked Mar 18 at 20:33

lidor718lidor718

125

$endgroup$

marked as duplicate by Mike Earnest, Shailesh, Cesareo, Eevee Trainer, Lee David Chung Lin Mar 19 at 1:46

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

add a comment | 

-1

$begingroup$

This question already has an answer here:

• 
  If any $15$ cars have $3$ with the same manufacturer, prove that any $100$ cars have $15$ with the same manufacturer.
  
  3 answers
  
  

I'm given the following problem:

In every group of 15 cars , there are 3 that were manufactured in the same country.
prove that in a group of 100 cars there are 15 cars that were manufactured in the same country.

does anyone have an idea of how to implement pigeonhole principle here? I'm kind of stuck because I can't assign cars to countries since I don't know how many countries are there..
Any help is appreciated . Thanks.

combinatorics pigeonhole-principle

share|cite|improve this question

edited Mar 18 at 21:02

Maria Mazur

49.4k1361124

asked Mar 18 at 20:33

lidor718lidor718

125

$endgroup$

marked as duplicate by Mike Earnest, Shailesh, Cesareo, Eevee Trainer, Lee David Chung Lin Mar 19 at 1:46

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

add a comment | 

$begingroup$

This question already has an answer here:

• 
  If any $15$ cars have $3$ with the same manufacturer, prove that any $100$ cars have $15$ with the same manufacturer.
  
  3 answers
  
  

I'm given the following problem:

In every group of 15 cars , there are 3 that were manufactured in the same country.
prove that in a group of 100 cars there are 15 cars that were manufactured in the same country.

does anyone have an idea of how to implement pigeonhole principle here? I'm kind of stuck because I can't assign cars to countries since I don't know how many countries are there..
Any help is appreciated . Thanks.

combinatorics pigeonhole-principle

share|cite|improve this question

edited Mar 18 at 21:02

Maria Mazur

49.4k1361124

asked Mar 18 at 20:33

lidor718lidor718

125

$endgroup$

share|cite|improve this question

edited Mar 18 at 21:02

Maria Mazur

49.4k1361124

asked Mar 18 at 20:33

lidor718lidor718

125

share|cite|improve this question

edited Mar 18 at 21:02

Maria Mazur

49.4k1361124

asked Mar 18 at 20:33

lidor718lidor718

125

edited Mar 18 at 21:02

Maria Mazur

49.4k1361124

edited Mar 18 at 21:02

Maria Mazur

49.4k1361124

asked Mar 18 at 20:33

lidor718lidor718

125

asked Mar 18 at 20:33

lidor718lidor718

125

2 Answers
2

active

oldest

votes

0

$begingroup$

First we argue that there cannot be more than $8$ countries. If there were, then we could choose $2$ cars of each of these $8$ countries and discard any of the chosen cars to receive a group of $15$ cars which does not have $3$ cars which were manufactured in the same country. So there are $leq 7$ countries. Thus by the pigeon-hole principle we have that for every group of $100$ cars at least $lceil100 / 7rceil = 15$ were manufactured in the same country.

Note that this argumentation fails if for some country there is only 1 car.

share|cite|improve this answer

answered Mar 18 at 20:46

araomisaraomis

4039

$endgroup$

add a comment | 

0

$begingroup$

Suppose each country manifactured at most $14$ cars and let $C_1,C_2,..C_n$ be al countries.

Then we have $$100=|C_1|+|C_2|+...+|C_n| leq 14n implies ngeq8$$

So we have at least $8$ countries. If we take from each country $2$ cars we have $16$ cars and no $3$ from the same country. A contradiction.

share|cite|improve this answer

answered Mar 18 at 20:59

Maria MazurMaria Mazur

49.4k1361124

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Note that the same caveat applies as in the other answer: If enough countries make only one car, this argument falls down.
  $endgroup$
  – Ingix
  Mar 18 at 21:32
  

  
  
  
  

add a comment | 

0

$begingroup$

First we argue that there cannot be more than $8$ countries. If there were, then we could choose $2$ cars of each of these $8$ countries and discard any of the chosen cars to receive a group of $15$ cars which does not have $3$ cars which were manufactured in the same country. So there are $leq 7$ countries. Thus by the pigeon-hole principle we have that for every group of $100$ cars at least $lceil100 / 7rceil = 15$ were manufactured in the same country.

Note that this argumentation fails if for some country there is only 1 car.

share|cite|improve this answer

answered Mar 18 at 20:46

araomisaraomis

4039

$endgroup$

add a comment | 

0

$begingroup$

First we argue that there cannot be more than $8$ countries. If there were, then we could choose $2$ cars of each of these $8$ countries and discard any of the chosen cars to receive a group of $15$ cars which does not have $3$ cars which were manufactured in the same country. So there are $leq 7$ countries. Thus by the pigeon-hole principle we have that for every group of $100$ cars at least $lceil100 / 7rceil = 15$ were manufactured in the same country.

Note that this argumentation fails if for some country there is only 1 car.

share|cite|improve this answer

answered Mar 18 at 20:46

araomisaraomis

4039

$endgroup$

add a comment | 

$begingroup$

First we argue that there cannot be more than $8$ countries. If there were, then we could choose $2$ cars of each of these $8$ countries and discard any of the chosen cars to receive a group of $15$ cars which does not have $3$ cars which were manufactured in the same country. So there are $leq 7$ countries. Thus by the pigeon-hole principle we have that for every group of $100$ cars at least $lceil100 / 7rceil = 15$ were manufactured in the same country.

Note that this argumentation fails if for some country there is only 1 car.

share|cite|improve this answer

answered Mar 18 at 20:46

araomisaraomis

4039

$endgroup$

share|cite|improve this answer

answered Mar 18 at 20:46

araomisaraomis

4039

answered Mar 18 at 20:46

araomisaraomis

4039

answered Mar 18 at 20:46

araomisaraomis

4039

0

$begingroup$

Suppose each country manifactured at most $14$ cars and let $C_1,C_2,..C_n$ be al countries.

Then we have $$100=|C_1|+|C_2|+...+|C_n| leq 14n implies ngeq8$$

So we have at least $8$ countries. If we take from each country $2$ cars we have $16$ cars and no $3$ from the same country. A contradiction.

share|cite|improve this answer

answered Mar 18 at 20:59

Maria MazurMaria Mazur

49.4k1361124

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Note that the same caveat applies as in the other answer: If enough countries make only one car, this argument falls down.
  $endgroup$
  – Ingix
  Mar 18 at 21:32
  

  
  
  
  

add a comment | 

0

$begingroup$

Suppose each country manifactured at most $14$ cars and let $C_1,C_2,..C_n$ be al countries.

Then we have $$100=|C_1|+|C_2|+...+|C_n| leq 14n implies ngeq8$$

So we have at least $8$ countries. If we take from each country $2$ cars we have $16$ cars and no $3$ from the same country. A contradiction.

share|cite|improve this answer

answered Mar 18 at 20:59

Maria MazurMaria Mazur

49.4k1361124

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Note that the same caveat applies as in the other answer: If enough countries make only one car, this argument falls down.
  $endgroup$
  – Ingix
  Mar 18 at 21:32
  

  
  
  
  

add a comment | 

$begingroup$

Suppose each country manifactured at most $14$ cars and let $C_1,C_2,..C_n$ be al countries.

Then we have $$100=|C_1|+|C_2|+...+|C_n| leq 14n implies ngeq8$$

So we have at least $8$ countries. If we take from each country $2$ cars we have $16$ cars and no $3$ from the same country. A contradiction.

share|cite|improve this answer

answered Mar 18 at 20:59

Maria MazurMaria Mazur

49.4k1361124

$endgroup$

share|cite|improve this answer

answered Mar 18 at 20:59

Maria MazurMaria Mazur

49.4k1361124

answered Mar 18 at 20:59

Maria MazurMaria Mazur

49.4k1361124

answered Mar 18 at 20:59

Maria MazurMaria Mazur

49.4k1361124