The pigeonhole principle - 30 pens in a drawer
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I need help with the following task. It needs to be solved using the pigeonhole principle.
There are 10 red, 8 blue, 8 purple and 4 yellow pens in a drawer. We pick them out, one by one, in the dark. What is the least number of pens that we need to pull out if we want to ensure that we have
a) at least 1 pen of each color
b) at least 6 blue pens?
I realize for example that if we want at lest 4 pens of each color, we need to pull out at least 11 ones, but I do not understand how to get one of each, or a specific number of a given color. I would solve this using statistics, but it has to be the principle.
combinatorics discrete-mathematics pigeonhole-principle
asked Nov 18 at 18:05
Tam
63
add a comment |
up vote
1
down vote
favorite
I need help with the following task. It needs to be solved using the pigeonhole principle.
There are 10 red, 8 blue, 8 purple and 4 yellow pens in a drawer. We pick them out, one by one, in the dark. What is the least number of pens that we need to pull out if we want to ensure that we have
a) at least 1 pen of each color
b) at least 6 blue pens?
I realize for example that if we want at lest 4 pens of each color, we need to pull out at least 11 ones, but I do not understand how to get one of each, or a specific number of a given color. I would solve this using statistics, but it has to be the principle.
combinatorics discrete-mathematics pigeonhole-principle
asked Nov 18 at 18:05
Tam
63
1
Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close with reading the comments.
– saulspatz
Nov 18 at 18:06
2
Think in terms of worst case scenarios. How many pens could you pull from the drawer before you finally obtain at least one pen of each color?
– N. F. Taussig
Nov 18 at 18:15
"I realize for example that if we want at least 4 pens of each color, we need to pull out at least 11 ones" does not look correct. Did you mean "if we want pens with at least 2 colors, we need to pull out at least 11 pens"?
– Henry
Nov 18 at 18:20
At most $27$ and $28$ respectively in the worst case.
– idea
Nov 18 at 18:20
Half-duplicate of math.stackexchange.com/questions/1612271/…
– Henry
Nov 18 at 18:22
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I need help with the following task. It needs to be solved using the pigeonhole principle.
There are 10 red, 8 blue, 8 purple and 4 yellow pens in a drawer. We pick them out, one by one, in the dark. What is the least number of pens that we need to pull out if we want to ensure that we have
a) at least 1 pen of each color
b) at least 6 blue pens?
I realize for example that if we want at lest 4 pens of each color, we need to pull out at least 11 ones, but I do not understand how to get one of each, or a specific number of a given color. I would solve this using statistics, but it has to be the principle.
combinatorics discrete-mathematics pigeonhole-principle
asked Nov 18 at 18:05
Tam
63
I need help with the following task. It needs to be solved using the pigeonhole principle.
There are 10 red, 8 blue, 8 purple and 4 yellow pens in a drawer. We pick them out, one by one, in the dark. What is the least number of pens that we need to pull out if we want to ensure that we have
a) at least 1 pen of each color
b) at least 6 blue pens?
I realize for example that if we want at lest 4 pens of each color, we need to pull out at least 11 ones, but I do not understand how to get one of each, or a specific number of a given color. I would solve this using statistics, but it has to be the principle.
combinatorics discrete-mathematics pigeonhole-principle
combinatorics discrete-mathematics pigeonhole-principle
asked Nov 18 at 18:05
Tam
63
asked Nov 18 at 18:05
Tam
63
edited Nov 18 at 18:16
asked Nov 18 at 18:05
Tam
63
asked Nov 18 at 18:05
Tam
63
asked Nov 18 at 18:05
Tam
63
63
1
Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close with reading the comments.
– saulspatz
Nov 18 at 18:06
2
Think in terms of worst case scenarios. How many pens could you pull from the drawer before you finally obtain at least one pen of each color?
– N. F. Taussig
Nov 18 at 18:15
"I realize for example that if we want at least 4 pens of each color, we need to pull out at least 11 ones" does not look correct. Did you mean "if we want pens with at least 2 colors, we need to pull out at least 11 pens"?
– Henry
Nov 18 at 18:20
At most $27$ and $28$ respectively in the worst case.
– idea
Nov 18 at 18:20
Half-duplicate of math.stackexchange.com/questions/1612271/…
– Henry
Nov 18 at 18:22
add a comment |
1
Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close with reading the comments.
– saulspatz
Nov 18 at 18:06
2
Think in terms of worst case scenarios. How many pens could you pull from the drawer before you finally obtain at least one pen of each color?
– N. F. Taussig
Nov 18 at 18:15
"I realize for example that if we want at least 4 pens of each color, we need to pull out at least 11 ones" does not look correct. Did you mean "if we want pens with at least 2 colors, we need to pull out at least 11 pens"?
– Henry
Nov 18 at 18:20
At most $27$ and $28$ respectively in the worst case.
– idea
Nov 18 at 18:20
Half-duplicate of math.stackexchange.com/questions/1612271/…
– Henry
Nov 18 at 18:22
1
1
Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close with reading the comments.
– saulspatz
Nov 18 at 18:06
Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close with reading the comments.
– saulspatz
Nov 18 at 18:06
2
2
Think in terms of worst case scenarios. How many pens could you pull from the drawer before you finally obtain at least one pen of each color?
– N. F. Taussig
Nov 18 at 18:15
Think in terms of worst case scenarios. How many pens could you pull from the drawer before you finally obtain at least one pen of each color?
– N. F. Taussig
Nov 18 at 18:15
"I realize for example that if we want at least 4 pens of each color, we need to pull out at least 11 ones" does not look correct. Did you mean "if we want pens with at least 2 colors, we need to pull out at least 11 pens"?
– Henry
Nov 18 at 18:20
"I realize for example that if we want at least 4 pens of each color, we need to pull out at least 11 ones" does not look correct. Did you mean "if we want pens with at least 2 colors, we need to pull out at least 11 pens"?
– Henry
Nov 18 at 18:20
At most $27$ and $28$ respectively in the worst case.
– idea
Nov 18 at 18:20
At most $27$ and $28$ respectively in the worst case.
– idea
Nov 18 at 18:20
Half-duplicate of math.stackexchange.com/questions/1612271/…
– Henry
Nov 18 at 18:22
Half-duplicate of math.stackexchange.com/questions/1612271/…
– Henry
Nov 18 at 18:22
add a comment |
1 Answer
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0
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Think logically.
Since you are drawing pens in the dark, it is very much possible that you require the maximum possible number of draws, i.e.
For $1^{st}$ case:
You might start off with Red, then finish off all Reds, Blues and Purples, i.e. $10+8+8=26$ draws in the worst case. Now, only left with Yellow, just draw one.
So, at most you may need $27$ draws.
While, if lucky enough, you may grab $4$ different colors in just $4$ draws.
So, to be sure, you will need at least $27$ draws.
For $2^{nd}$ case:
Finish drawing all other colors, contributing to $10+8+4=22$ draws. Now, left with only Blues, just draw $6$, so at most $22+6=28$ draws in the worst-case.
Again, in the rarest case, you may need just $6$ draws; though its probability is quite less.
So, to be sure, you will need at least $28$ draws.
Try to convince yourself!
answered Nov 18 at 18:29
idea
1,96231024
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Think logically.
Since you are drawing pens in the dark, it is very much possible that you require the maximum possible number of draws, i.e.
For $1^{st}$ case:
You might start off with Red, then finish off all Reds, Blues and Purples, i.e. $10+8+8=26$ draws in the worst case. Now, only left with Yellow, just draw one.
So, at most you may need $27$ draws.
While, if lucky enough, you may grab $4$ different colors in just $4$ draws.
So, to be sure, you will need at least $27$ draws.
For $2^{nd}$ case:
Finish drawing all other colors, contributing to $10+8+4=22$ draws. Now, left with only Blues, just draw $6$, so at most $22+6=28$ draws in the worst-case.
Again, in the rarest case, you may need just $6$ draws; though its probability is quite less.
So, to be sure, you will need at least $28$ draws.
Try to convince yourself!
answered Nov 18 at 18:29
idea
1,96231024
add a comment |
up vote
0
down vote
Think logically.
Since you are drawing pens in the dark, it is very much possible that you require the maximum possible number of draws, i.e.
For $1^{st}$ case:
You might start off with Red, then finish off all Reds, Blues and Purples, i.e. $10+8+8=26$ draws in the worst case. Now, only left with Yellow, just draw one.
So, at most you may need $27$ draws.
While, if lucky enough, you may grab $4$ different colors in just $4$ draws.
So, to be sure, you will need at least $27$ draws.
For $2^{nd}$ case:
Finish drawing all other colors, contributing to $10+8+4=22$ draws. Now, left with only Blues, just draw $6$, so at most $22+6=28$ draws in the worst-case.
Again, in the rarest case, you may need just $6$ draws; though its probability is quite less.
So, to be sure, you will need at least $28$ draws.
Try to convince yourself!
answered Nov 18 at 18:29
idea
1,96231024
add a comment |
up vote
0
down vote
up vote
0
down vote
Think logically.
Since you are drawing pens in the dark, it is very much possible that you require the maximum possible number of draws, i.e.
For $1^{st}$ case:
You might start off with Red, then finish off all Reds, Blues and Purples, i.e. $10+8+8=26$ draws in the worst case. Now, only left with Yellow, just draw one.
So, at most you may need $27$ draws.
While, if lucky enough, you may grab $4$ different colors in just $4$ draws.
So, to be sure, you will need at least $27$ draws.
For $2^{nd}$ case:
Finish drawing all other colors, contributing to $10+8+4=22$ draws. Now, left with only Blues, just draw $6$, so at most $22+6=28$ draws in the worst-case.
Again, in the rarest case, you may need just $6$ draws; though its probability is quite less.
So, to be sure, you will need at least $28$ draws.
Try to convince yourself!
answered Nov 18 at 18:29
idea
1,96231024
Think logically.
Since you are drawing pens in the dark, it is very much possible that you require the maximum possible number of draws, i.e.
For $1^{st}$ case:
You might start off with Red, then finish off all Reds, Blues and Purples, i.e. $10+8+8=26$ draws in the worst case. Now, only left with Yellow, just draw one.
So, at most you may need $27$ draws.
While, if lucky enough, you may grab $4$ different colors in just $4$ draws.
So, to be sure, you will need at least $27$ draws.
For $2^{nd}$ case:
Finish drawing all other colors, contributing to $10+8+4=22$ draws. Now, left with only Blues, just draw $6$, so at most $22+6=28$ draws in the worst-case.
Again, in the rarest case, you may need just $6$ draws; though its probability is quite less.
So, to be sure, you will need at least $28$ draws.
Try to convince yourself!
answered Nov 18 at 18:29
idea
1,96231024
edited Nov 18 at 18:35
answered Nov 18 at 18:29
idea
1,96231024
answered Nov 18 at 18:29
idea
1,96231024
answered Nov 18 at 18:29
idea
1,96231024
1,96231024
add a comment |
add a comment |
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Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close with reading the comments.
– saulspatz
Nov 18 at 18:06
2
Think in terms of worst case scenarios. How many pens could you pull from the drawer before you finally obtain at least one pen of each color?
– N. F. Taussig
Nov 18 at 18:15
"I realize for example that if we want at least 4 pens of each color, we need to pull out at least 11 ones" does not look correct. Did you mean "if we want pens with at least 2 colors, we need to pull out at least 11 pens"?
– Henry
Nov 18 at 18:20
At most $27$ and $28$ respectively in the worst case.
– idea
Nov 18 at 18:20
Half-duplicate of math.stackexchange.com/questions/1612271/…
– Henry
Nov 18 at 18:22