Pigeonhole Principle Proof Generalized
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An airport sees 1500 takeoffs per day. Prove that there are two planes that leave within a minute of each other.
All I can get started with is finding the total minutes in a day- 1440min. I understand that I have to fit 1500 planes into 1440 boxes however I do not know how to use this to prove that at least 2 planes leave within a minute of each other. Kindly help.
probability combinatorics discrete-mathematics pigeonhole-principle
asked Dec 9 '18 at 13:03
MunchiesOatsMunchiesOats
323
$endgroup$
add a comment |
$begingroup$
An airport sees 1500 takeoffs per day. Prove that there are two planes that leave within a minute of each other.
All I can get started with is finding the total minutes in a day- 1440min. I understand that I have to fit 1500 planes into 1440 boxes however I do not know how to use this to prove that at least 2 planes leave within a minute of each other. Kindly help.
probability combinatorics discrete-mathematics pigeonhole-principle
asked Dec 9 '18 at 13:03
MunchiesOatsMunchiesOats
323
$endgroup$
2
$begingroup$
assume a plane can only leave every minute, only 1440 planes can leave. If there are 1500 flights, then at least 60 of them must leave within same minutes.
$endgroup$
– user29418
Dec 9 '18 at 13:06
1
$begingroup$
@user29418 It is not necessary that atleast 60 of them must leave within the same minutes. You can say that 60 pairs are there must leave within same minute
$endgroup$
– user408906
Dec 9 '18 at 13:10
$begingroup$
you're right I messed up the wording
$endgroup$
– user29418
Dec 10 '18 at 1:18
add a comment |
$begingroup$
An airport sees 1500 takeoffs per day. Prove that there are two planes that leave within a minute of each other.
All I can get started with is finding the total minutes in a day- 1440min. I understand that I have to fit 1500 planes into 1440 boxes however I do not know how to use this to prove that at least 2 planes leave within a minute of each other. Kindly help.
probability combinatorics discrete-mathematics pigeonhole-principle
asked Dec 9 '18 at 13:03
MunchiesOatsMunchiesOats
323
$endgroup$
An airport sees 1500 takeoffs per day. Prove that there are two planes that leave within a minute of each other.
All I can get started with is finding the total minutes in a day- 1440min. I understand that I have to fit 1500 planes into 1440 boxes however I do not know how to use this to prove that at least 2 planes leave within a minute of each other. Kindly help.
probability combinatorics discrete-mathematics pigeonhole-principle
probability combinatorics discrete-mathematics pigeonhole-principle
asked Dec 9 '18 at 13:03
MunchiesOatsMunchiesOats
323
asked Dec 9 '18 at 13:03
MunchiesOatsMunchiesOats
323
asked Dec 9 '18 at 13:03
MunchiesOatsMunchiesOats
323
asked Dec 9 '18 at 13:03
MunchiesOatsMunchiesOats
323
asked Dec 9 '18 at 13:03
MunchiesOatsMunchiesOats
323
323
2
$begingroup$
assume a plane can only leave every minute, only 1440 planes can leave. If there are 1500 flights, then at least 60 of them must leave within same minutes.
$endgroup$
– user29418
Dec 9 '18 at 13:06
1
$begingroup$
@user29418 It is not necessary that atleast 60 of them must leave within the same minutes. You can say that 60 pairs are there must leave within same minute
$endgroup$
– user408906
Dec 9 '18 at 13:10
$begingroup$
you're right I messed up the wording
$endgroup$
– user29418
Dec 10 '18 at 1:18
add a comment |
2
$begingroup$
assume a plane can only leave every minute, only 1440 planes can leave. If there are 1500 flights, then at least 60 of them must leave within same minutes.
$endgroup$
– user29418
Dec 9 '18 at 13:06
1
$begingroup$
@user29418 It is not necessary that atleast 60 of them must leave within the same minutes. You can say that 60 pairs are there must leave within same minute
$endgroup$
– user408906
Dec 9 '18 at 13:10
$begingroup$
you're right I messed up the wording
$endgroup$
– user29418
Dec 10 '18 at 1:18
2
2
$begingroup$
assume a plane can only leave every minute, only 1440 planes can leave. If there are 1500 flights, then at least 60 of them must leave within same minutes.
$endgroup$
– user29418
Dec 9 '18 at 13:06
$begingroup$
assume a plane can only leave every minute, only 1440 planes can leave. If there are 1500 flights, then at least 60 of them must leave within same minutes.
$endgroup$
– user29418
Dec 9 '18 at 13:06
1
1
$begingroup$
@user29418 It is not necessary that atleast 60 of them must leave within the same minutes. You can say that 60 pairs are there must leave within same minute
$endgroup$
– user408906
Dec 9 '18 at 13:10
$begingroup$
@user29418 It is not necessary that atleast 60 of them must leave within the same minutes. You can say that 60 pairs are there must leave within same minute
$endgroup$
– user408906
Dec 9 '18 at 13:10
$begingroup$
you're right I messed up the wording
$endgroup$
– user29418
Dec 10 '18 at 1:18
$begingroup$
you're right I messed up the wording
$endgroup$
– user29418
Dec 10 '18 at 1:18
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The Pigeonhole Principle seems to be the perfect approach.
Let the 1440 minutes be the pigeonholes and the 1500 takeoffs be the pigeons. It follows that $$text{At least } lceilfrac{1500}{1440}rceil=2text{ planes take off in the same minute}$$
answered Dec 9 '18 at 13:17
Dr. MathvaDr. Mathva
1,059316
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add a comment |
$begingroup$
Imagine the day as divided into $1440$ one-minute intervals. Those will be your pigeonholes. The $1500$ planes are your (high-tech) pigeons. Think of a plane (= pigeon) leaving within a particular minute (= pigeonhole) as putting that pigeon in that hole. The pigeonhole principle says there must be two pigeons in a single hole. Given the nature of your pigeons, holes, and "pigeon being in a hole", this means that there must be two planes leaving in the same one-minute interval.
In general, when applying the pigeonhole principle to problems like this, you should look for ways to interpret "pigeon", "hole", and "pigeon being in a hole" in ways relevant to your problem, so that the conclusion of the pigeonhole principle, "at least two pigeons in the same hole" gets interpreted as the conclusion you want.
answered Dec 9 '18 at 13:18
Andreas BlassAndreas Blass
49.6k451108
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add a comment |
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2 Answers
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active
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2 Answers
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active
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active
oldest
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$begingroup$
The Pigeonhole Principle seems to be the perfect approach.
Let the 1440 minutes be the pigeonholes and the 1500 takeoffs be the pigeons. It follows that $$text{At least } lceilfrac{1500}{1440}rceil=2text{ planes take off in the same minute}$$
answered Dec 9 '18 at 13:17
Dr. MathvaDr. Mathva
1,059316
$endgroup$
add a comment |
$begingroup$
The Pigeonhole Principle seems to be the perfect approach.
Let the 1440 minutes be the pigeonholes and the 1500 takeoffs be the pigeons. It follows that $$text{At least } lceilfrac{1500}{1440}rceil=2text{ planes take off in the same minute}$$
answered Dec 9 '18 at 13:17
Dr. MathvaDr. Mathva
1,059316
$endgroup$
add a comment |
$begingroup$
The Pigeonhole Principle seems to be the perfect approach.
Let the 1440 minutes be the pigeonholes and the 1500 takeoffs be the pigeons. It follows that $$text{At least } lceilfrac{1500}{1440}rceil=2text{ planes take off in the same minute}$$
answered Dec 9 '18 at 13:17
Dr. MathvaDr. Mathva
1,059316
$endgroup$
The Pigeonhole Principle seems to be the perfect approach.
Let the 1440 minutes be the pigeonholes and the 1500 takeoffs be the pigeons. It follows that $$text{At least } lceilfrac{1500}{1440}rceil=2text{ planes take off in the same minute}$$
answered Dec 9 '18 at 13:17
Dr. MathvaDr. Mathva
1,059316
answered Dec 9 '18 at 13:17
Dr. MathvaDr. Mathva
1,059316
answered Dec 9 '18 at 13:17
Dr. MathvaDr. Mathva
1,059316
answered Dec 9 '18 at 13:17
Dr. MathvaDr. Mathva
1,059316
1,059316
add a comment |
add a comment |
$begingroup$
Imagine the day as divided into $1440$ one-minute intervals. Those will be your pigeonholes. The $1500$ planes are your (high-tech) pigeons. Think of a plane (= pigeon) leaving within a particular minute (= pigeonhole) as putting that pigeon in that hole. The pigeonhole principle says there must be two pigeons in a single hole. Given the nature of your pigeons, holes, and "pigeon being in a hole", this means that there must be two planes leaving in the same one-minute interval.
In general, when applying the pigeonhole principle to problems like this, you should look for ways to interpret "pigeon", "hole", and "pigeon being in a hole" in ways relevant to your problem, so that the conclusion of the pigeonhole principle, "at least two pigeons in the same hole" gets interpreted as the conclusion you want.
answered Dec 9 '18 at 13:18
Andreas BlassAndreas Blass
49.6k451108
$endgroup$
add a comment |
$begingroup$
Imagine the day as divided into $1440$ one-minute intervals. Those will be your pigeonholes. The $1500$ planes are your (high-tech) pigeons. Think of a plane (= pigeon) leaving within a particular minute (= pigeonhole) as putting that pigeon in that hole. The pigeonhole principle says there must be two pigeons in a single hole. Given the nature of your pigeons, holes, and "pigeon being in a hole", this means that there must be two planes leaving in the same one-minute interval.
In general, when applying the pigeonhole principle to problems like this, you should look for ways to interpret "pigeon", "hole", and "pigeon being in a hole" in ways relevant to your problem, so that the conclusion of the pigeonhole principle, "at least two pigeons in the same hole" gets interpreted as the conclusion you want.
answered Dec 9 '18 at 13:18
Andreas BlassAndreas Blass
49.6k451108
$endgroup$
add a comment |
$begingroup$
Imagine the day as divided into $1440$ one-minute intervals. Those will be your pigeonholes. The $1500$ planes are your (high-tech) pigeons. Think of a plane (= pigeon) leaving within a particular minute (= pigeonhole) as putting that pigeon in that hole. The pigeonhole principle says there must be two pigeons in a single hole. Given the nature of your pigeons, holes, and "pigeon being in a hole", this means that there must be two planes leaving in the same one-minute interval.
In general, when applying the pigeonhole principle to problems like this, you should look for ways to interpret "pigeon", "hole", and "pigeon being in a hole" in ways relevant to your problem, so that the conclusion of the pigeonhole principle, "at least two pigeons in the same hole" gets interpreted as the conclusion you want.
answered Dec 9 '18 at 13:18
Andreas BlassAndreas Blass
49.6k451108
$endgroup$
Imagine the day as divided into $1440$ one-minute intervals. Those will be your pigeonholes. The $1500$ planes are your (high-tech) pigeons. Think of a plane (= pigeon) leaving within a particular minute (= pigeonhole) as putting that pigeon in that hole. The pigeonhole principle says there must be two pigeons in a single hole. Given the nature of your pigeons, holes, and "pigeon being in a hole", this means that there must be two planes leaving in the same one-minute interval.
In general, when applying the pigeonhole principle to problems like this, you should look for ways to interpret "pigeon", "hole", and "pigeon being in a hole" in ways relevant to your problem, so that the conclusion of the pigeonhole principle, "at least two pigeons in the same hole" gets interpreted as the conclusion you want.
answered Dec 9 '18 at 13:18
Andreas BlassAndreas Blass
49.6k451108
answered Dec 9 '18 at 13:18
Andreas BlassAndreas Blass
49.6k451108
answered Dec 9 '18 at 13:18
Andreas BlassAndreas Blass
49.6k451108
answered Dec 9 '18 at 13:18
Andreas BlassAndreas Blass
49.6k451108
49.6k451108
add a comment |
add a comment |
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$begingroup$
assume a plane can only leave every minute, only 1440 planes can leave. If there are 1500 flights, then at least 60 of them must leave within same minutes.
$endgroup$
– user29418
Dec 9 '18 at 13:06
1
$begingroup$
@user29418 It is not necessary that atleast 60 of them must leave within the same minutes. You can say that 60 pairs are there must leave within same minute
$endgroup$
– user408906
Dec 9 '18 at 13:10
$begingroup$
you're right I messed up the wording
$endgroup$
– user29418
Dec 10 '18 at 1:18