Pigeon Hole explanation Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Min Number of Values from 1,2,…,9 Such that diff of 2 picked values is 5Pigeon Hole Principle AlgorithmThe Probabilistic Pigeon Hole PrinciplePigeon Hole Priciple and Genaralized Pigeon Hole Principle QuestionPigeonhole problem - Can solve it but can't model how it works…Tips on identifying pigeon and pigeonholeleast number of items required to satisfy one of three given conditions?Not quite understanding parts of Pigeon Hole Principle GeneralizationK-subsets, counting, and the pigeon hole principlePigeon hole principle proof writing
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Pigeon Hole explanation
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Min Number of Values from 1,2,…,9 Such that diff of 2 picked values is 5Pigeon Hole Principle AlgorithmThe Probabilistic Pigeon Hole PrinciplePigeon Hole Priciple and Genaralized Pigeon Hole Principle QuestionPigeonhole problem - Can solve it but can't model how it works…Tips on identifying pigeon and pigeonholeleast number of items required to satisfy one of three given conditions?Not quite understanding parts of Pigeon Hole Principle GeneralizationK-subsets, counting, and the pigeon hole principlePigeon hole principle proof writing
$begingroup$
I understand that the pigeonhole principle is supposedly a quite simple concept. However could you please explain to me the reasoning of how you reach this answer. Thank you.
Question: A basket cannot contain more than $24$ apples. What is the minimum amount of baskets you must have, to ensure you have at least $5$ baskets with the same number of apples in them (all baskets have at least $1$ apple contained within).
Answer of this question being $97$ baskets.
pigeonhole-principle
$endgroup$
add a comment |
$begingroup$
I understand that the pigeonhole principle is supposedly a quite simple concept. However could you please explain to me the reasoning of how you reach this answer. Thank you.
Question: A basket cannot contain more than $24$ apples. What is the minimum amount of baskets you must have, to ensure you have at least $5$ baskets with the same number of apples in them (all baskets have at least $1$ apple contained within).
Answer of this question being $97$ baskets.
pigeonhole-principle
$endgroup$
add a comment |
$begingroup$
I understand that the pigeonhole principle is supposedly a quite simple concept. However could you please explain to me the reasoning of how you reach this answer. Thank you.
Question: A basket cannot contain more than $24$ apples. What is the minimum amount of baskets you must have, to ensure you have at least $5$ baskets with the same number of apples in them (all baskets have at least $1$ apple contained within).
Answer of this question being $97$ baskets.
pigeonhole-principle
$endgroup$
I understand that the pigeonhole principle is supposedly a quite simple concept. However could you please explain to me the reasoning of how you reach this answer. Thank you.
Question: A basket cannot contain more than $24$ apples. What is the minimum amount of baskets you must have, to ensure you have at least $5$ baskets with the same number of apples in them (all baskets have at least $1$ apple contained within).
Answer of this question being $97$ baskets.
pigeonhole-principle
pigeonhole-principle
edited 19 hours ago
Peter
49.2k1240138
49.2k1240138
asked 19 hours ago
LaykenLayken
324
324
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
The basic idea of the pigeonhole principle is trivial , but the application can be much more difficult.
Main idea : If we distribute $n+1$ pigeons among $n$ cages, at least one cage must have more than one pigeon.
The problem here :
There are $24$ possibilities for the number of apples in a basket.
Therefore, $96$ baskets cannot be enough because every number from $1$ to $24$ can appear exactly four times.
But if we add another basket, it is not possible anymore that all the numbers appear at most $4$ times because then, at most $96$ baskets would be possible.
$endgroup$
add a comment |
$begingroup$
Why $96$ is not enought:
For every number $x$ between $1$ and $24$ you take $4$ baskets with $x$ apples inside. Then you have $4cdot 24 = 96$ baskets. By construction, there is no number such that $5$ baskets have this number of apples.
On the other hand if you have $97$ baskets and assume there are maximum $4$ baskets with the same number of apples. Then again the number of baskets is limited to $4cdot 24$ which is less than your number of baskets. So you have a contradiction. Hence, there is no configuration with max $4$ baskets with the same number of apples for $97$ baskets.
$endgroup$
add a comment |
$begingroup$
Since a basket cannot contain more than $24$ apples, we consider a basket to be a collection of $24$ pigeonholes. Hence if we have $4$ baskets, there are altogether $24 times 4 = 96$ pigeonholes. Indeed $96$ pigeonholes can contain $96$ pigeons (apples). But if we have $97$ pigeons, by pigeonhole principle, at least one of the pigeonhole must have two pigeons, which is not allowed (the $4$ baskets are full, so we need an extra one). Hence $97$ is the answer.
$endgroup$
add a comment |
$begingroup$
The number of apples in a basket is between 1 and 24. That's 24 different values.
Suppose that in our collection of baskets, none of these numbers occurs at least five times.
This means that in our collection of baskets each of these numbers occurs at most four times.
But then we can have at most 24 times 4 baskets.
Now, $24 times 4 = 96$. So if we have more than that number of baskets, that is if we have at least 97 baskets, then one of those numbers between 1 and 24 must occur at least five times, that is then we have at least five baskets with the same number of apples.
$endgroup$
$begingroup$
Please don't use $24 * 4 = 96$, it looks bad ... Use $24 times 4$ or $24 cdot 4$.
$endgroup$
– L. F.
18 hours ago
add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
The basic idea of the pigeonhole principle is trivial , but the application can be much more difficult.
Main idea : If we distribute $n+1$ pigeons among $n$ cages, at least one cage must have more than one pigeon.
The problem here :
There are $24$ possibilities for the number of apples in a basket.
Therefore, $96$ baskets cannot be enough because every number from $1$ to $24$ can appear exactly four times.
But if we add another basket, it is not possible anymore that all the numbers appear at most $4$ times because then, at most $96$ baskets would be possible.
$endgroup$
add a comment |
$begingroup$
The basic idea of the pigeonhole principle is trivial , but the application can be much more difficult.
Main idea : If we distribute $n+1$ pigeons among $n$ cages, at least one cage must have more than one pigeon.
The problem here :
There are $24$ possibilities for the number of apples in a basket.
Therefore, $96$ baskets cannot be enough because every number from $1$ to $24$ can appear exactly four times.
But if we add another basket, it is not possible anymore that all the numbers appear at most $4$ times because then, at most $96$ baskets would be possible.
$endgroup$
add a comment |
$begingroup$
The basic idea of the pigeonhole principle is trivial , but the application can be much more difficult.
Main idea : If we distribute $n+1$ pigeons among $n$ cages, at least one cage must have more than one pigeon.
The problem here :
There are $24$ possibilities for the number of apples in a basket.
Therefore, $96$ baskets cannot be enough because every number from $1$ to $24$ can appear exactly four times.
But if we add another basket, it is not possible anymore that all the numbers appear at most $4$ times because then, at most $96$ baskets would be possible.
$endgroup$
The basic idea of the pigeonhole principle is trivial , but the application can be much more difficult.
Main idea : If we distribute $n+1$ pigeons among $n$ cages, at least one cage must have more than one pigeon.
The problem here :
There are $24$ possibilities for the number of apples in a basket.
Therefore, $96$ baskets cannot be enough because every number from $1$ to $24$ can appear exactly four times.
But if we add another basket, it is not possible anymore that all the numbers appear at most $4$ times because then, at most $96$ baskets would be possible.
answered 19 hours ago
PeterPeter
49.2k1240138
49.2k1240138
add a comment |
add a comment |
$begingroup$
Why $96$ is not enought:
For every number $x$ between $1$ and $24$ you take $4$ baskets with $x$ apples inside. Then you have $4cdot 24 = 96$ baskets. By construction, there is no number such that $5$ baskets have this number of apples.
On the other hand if you have $97$ baskets and assume there are maximum $4$ baskets with the same number of apples. Then again the number of baskets is limited to $4cdot 24$ which is less than your number of baskets. So you have a contradiction. Hence, there is no configuration with max $4$ baskets with the same number of apples for $97$ baskets.
$endgroup$
add a comment |
$begingroup$
Why $96$ is not enought:
For every number $x$ between $1$ and $24$ you take $4$ baskets with $x$ apples inside. Then you have $4cdot 24 = 96$ baskets. By construction, there is no number such that $5$ baskets have this number of apples.
On the other hand if you have $97$ baskets and assume there are maximum $4$ baskets with the same number of apples. Then again the number of baskets is limited to $4cdot 24$ which is less than your number of baskets. So you have a contradiction. Hence, there is no configuration with max $4$ baskets with the same number of apples for $97$ baskets.
$endgroup$
add a comment |
$begingroup$
Why $96$ is not enought:
For every number $x$ between $1$ and $24$ you take $4$ baskets with $x$ apples inside. Then you have $4cdot 24 = 96$ baskets. By construction, there is no number such that $5$ baskets have this number of apples.
On the other hand if you have $97$ baskets and assume there are maximum $4$ baskets with the same number of apples. Then again the number of baskets is limited to $4cdot 24$ which is less than your number of baskets. So you have a contradiction. Hence, there is no configuration with max $4$ baskets with the same number of apples for $97$ baskets.
$endgroup$
Why $96$ is not enought:
For every number $x$ between $1$ and $24$ you take $4$ baskets with $x$ apples inside. Then you have $4cdot 24 = 96$ baskets. By construction, there is no number such that $5$ baskets have this number of apples.
On the other hand if you have $97$ baskets and assume there are maximum $4$ baskets with the same number of apples. Then again the number of baskets is limited to $4cdot 24$ which is less than your number of baskets. So you have a contradiction. Hence, there is no configuration with max $4$ baskets with the same number of apples for $97$ baskets.
answered 19 hours ago
Nathanael SkrepekNathanael Skrepek
1,7921615
1,7921615
add a comment |
add a comment |
$begingroup$
Since a basket cannot contain more than $24$ apples, we consider a basket to be a collection of $24$ pigeonholes. Hence if we have $4$ baskets, there are altogether $24 times 4 = 96$ pigeonholes. Indeed $96$ pigeonholes can contain $96$ pigeons (apples). But if we have $97$ pigeons, by pigeonhole principle, at least one of the pigeonhole must have two pigeons, which is not allowed (the $4$ baskets are full, so we need an extra one). Hence $97$ is the answer.
$endgroup$
add a comment |
$begingroup$
Since a basket cannot contain more than $24$ apples, we consider a basket to be a collection of $24$ pigeonholes. Hence if we have $4$ baskets, there are altogether $24 times 4 = 96$ pigeonholes. Indeed $96$ pigeonholes can contain $96$ pigeons (apples). But if we have $97$ pigeons, by pigeonhole principle, at least one of the pigeonhole must have two pigeons, which is not allowed (the $4$ baskets are full, so we need an extra one). Hence $97$ is the answer.
$endgroup$
add a comment |
$begingroup$
Since a basket cannot contain more than $24$ apples, we consider a basket to be a collection of $24$ pigeonholes. Hence if we have $4$ baskets, there are altogether $24 times 4 = 96$ pigeonholes. Indeed $96$ pigeonholes can contain $96$ pigeons (apples). But if we have $97$ pigeons, by pigeonhole principle, at least one of the pigeonhole must have two pigeons, which is not allowed (the $4$ baskets are full, so we need an extra one). Hence $97$ is the answer.
$endgroup$
Since a basket cannot contain more than $24$ apples, we consider a basket to be a collection of $24$ pigeonholes. Hence if we have $4$ baskets, there are altogether $24 times 4 = 96$ pigeonholes. Indeed $96$ pigeonholes can contain $96$ pigeons (apples). But if we have $97$ pigeons, by pigeonhole principle, at least one of the pigeonhole must have two pigeons, which is not allowed (the $4$ baskets are full, so we need an extra one). Hence $97$ is the answer.
answered 19 hours ago
tonychow0929tonychow0929
42137
42137
add a comment |
add a comment |
$begingroup$
The number of apples in a basket is between 1 and 24. That's 24 different values.
Suppose that in our collection of baskets, none of these numbers occurs at least five times.
This means that in our collection of baskets each of these numbers occurs at most four times.
But then we can have at most 24 times 4 baskets.
Now, $24 times 4 = 96$. So if we have more than that number of baskets, that is if we have at least 97 baskets, then one of those numbers between 1 and 24 must occur at least five times, that is then we have at least five baskets with the same number of apples.
$endgroup$
$begingroup$
Please don't use $24 * 4 = 96$, it looks bad ... Use $24 times 4$ or $24 cdot 4$.
$endgroup$
– L. F.
18 hours ago
add a comment |
$begingroup$
The number of apples in a basket is between 1 and 24. That's 24 different values.
Suppose that in our collection of baskets, none of these numbers occurs at least five times.
This means that in our collection of baskets each of these numbers occurs at most four times.
But then we can have at most 24 times 4 baskets.
Now, $24 times 4 = 96$. So if we have more than that number of baskets, that is if we have at least 97 baskets, then one of those numbers between 1 and 24 must occur at least five times, that is then we have at least five baskets with the same number of apples.
$endgroup$
$begingroup$
Please don't use $24 * 4 = 96$, it looks bad ... Use $24 times 4$ or $24 cdot 4$.
$endgroup$
– L. F.
18 hours ago
add a comment |
$begingroup$
The number of apples in a basket is between 1 and 24. That's 24 different values.
Suppose that in our collection of baskets, none of these numbers occurs at least five times.
This means that in our collection of baskets each of these numbers occurs at most four times.
But then we can have at most 24 times 4 baskets.
Now, $24 times 4 = 96$. So if we have more than that number of baskets, that is if we have at least 97 baskets, then one of those numbers between 1 and 24 must occur at least five times, that is then we have at least five baskets with the same number of apples.
$endgroup$
The number of apples in a basket is between 1 and 24. That's 24 different values.
Suppose that in our collection of baskets, none of these numbers occurs at least five times.
This means that in our collection of baskets each of these numbers occurs at most four times.
But then we can have at most 24 times 4 baskets.
Now, $24 times 4 = 96$. So if we have more than that number of baskets, that is if we have at least 97 baskets, then one of those numbers between 1 and 24 must occur at least five times, that is then we have at least five baskets with the same number of apples.
edited 5 hours ago
answered 19 hours ago
jflippjflipp
3,7711711
3,7711711
$begingroup$
Please don't use $24 * 4 = 96$, it looks bad ... Use $24 times 4$ or $24 cdot 4$.
$endgroup$
– L. F.
18 hours ago
add a comment |
$begingroup$
Please don't use $24 * 4 = 96$, it looks bad ... Use $24 times 4$ or $24 cdot 4$.
$endgroup$
– L. F.
18 hours ago
$begingroup$
Please don't use $24 * 4 = 96$, it looks bad ... Use $24 times 4$ or $24 cdot 4$.
$endgroup$
– L. F.
18 hours ago
$begingroup$
Please don't use $24 * 4 = 96$, it looks bad ... Use $24 times 4$ or $24 cdot 4$.
$endgroup$
– L. F.
18 hours ago
add a comment |
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