Pigeonhole Principle Question - Selecting some items from a box that contains fifty items with different...












1












$begingroup$


Please help me understand this question. I got the answer 101, however it says the solution to this question is 11. Perhaps it was a typo?



Suppose you select some items from a box that contains fifty items, where there are ten each of the colors blue, red, green, yellow, and purple. How many items must you select to make sure that you get at
least three of the same color?



As requested, I will express the problem and my thoughts. Note that ceiling division is used in this problem. I will denote ceiling division with the symbols [x], and I will use <= for less than or equal to.



3 <= [X/50]



2 < X/50



100 < X



101 <= X



This is how my discrete structures professor requested how I complete this problem.






pigeonhole-principle







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The answer is correct. Note that 101 doesn’t seem very reasonable considering that there are only 50 items to begin with.
    $endgroup$
    – platty
    Dec 1 '18 at 1:28










  • $begingroup$
    Why not tell everybody your thoughts about this question? A single answer is not helpful to you. People would be glad to check your reasoning and tell you which part is wrong. Please add your thoughts in the post, not in the comment section.
    $endgroup$
    – xbh
    Dec 1 '18 at 1:36












  • $begingroup$
    It's $2 < frac{X}{5}$ and $3 leq leftlceil frac{X}{5} rightrceil$.
    $endgroup$
    – Fabio Somenzi
    Dec 1 '18 at 1:51












  • $begingroup$
    Please edit the title of the question into something more descriptive.
    $endgroup$
    – Shaun
    Dec 1 '18 at 2:06










  • $begingroup$
    I think you read the question wrong: there are 10 items for each color, i.e. 10 blue, 10 red, 10 green, 10 yellow, and 10 purple.
    $endgroup$
    – xbh
    Dec 1 '18 at 2:56
















1












$begingroup$


Please help me understand this question. I got the answer 101, however it says the solution to this question is 11. Perhaps it was a typo?



Suppose you select some items from a box that contains fifty items, where there are ten each of the colors blue, red, green, yellow, and purple. How many items must you select to make sure that you get at
least three of the same color?



As requested, I will express the problem and my thoughts. Note that ceiling division is used in this problem. I will denote ceiling division with the symbols [x], and I will use <= for less than or equal to.



3 <= [X/50]



2 < X/50



100 < X



101 <= X



This is how my discrete structures professor requested how I complete this problem.






pigeonhole-principle







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The answer is correct. Note that 101 doesn’t seem very reasonable considering that there are only 50 items to begin with.
    $endgroup$
    – platty
    Dec 1 '18 at 1:28










  • $begingroup$
    Why not tell everybody your thoughts about this question? A single answer is not helpful to you. People would be glad to check your reasoning and tell you which part is wrong. Please add your thoughts in the post, not in the comment section.
    $endgroup$
    – xbh
    Dec 1 '18 at 1:36












  • $begingroup$
    It's $2 < frac{X}{5}$ and $3 leq leftlceil frac{X}{5} rightrceil$.
    $endgroup$
    – Fabio Somenzi
    Dec 1 '18 at 1:51












  • $begingroup$
    Please edit the title of the question into something more descriptive.
    $endgroup$
    – Shaun
    Dec 1 '18 at 2:06










  • $begingroup$
    I think you read the question wrong: there are 10 items for each color, i.e. 10 blue, 10 red, 10 green, 10 yellow, and 10 purple.
    $endgroup$
    – xbh
    Dec 1 '18 at 2:56














1












1








1


1



$begingroup$


Please help me understand this question. I got the answer 101, however it says the solution to this question is 11. Perhaps it was a typo?



Suppose you select some items from a box that contains fifty items, where there are ten each of the colors blue, red, green, yellow, and purple. How many items must you select to make sure that you get at
least three of the same color?



As requested, I will express the problem and my thoughts. Note that ceiling division is used in this problem. I will denote ceiling division with the symbols [x], and I will use <= for less than or equal to.



3 <= [X/50]



2 < X/50



100 < X



101 <= X



This is how my discrete structures professor requested how I complete this problem.






pigeonhole-principle







share|cite|improve this question











$endgroup$




Please help me understand this question. I got the answer 101, however it says the solution to this question is 11. Perhaps it was a typo?



Suppose you select some items from a box that contains fifty items, where there are ten each of the colors blue, red, green, yellow, and purple. How many items must you select to make sure that you get at
least three of the same color?



As requested, I will express the problem and my thoughts. Note that ceiling division is used in this problem. I will denote ceiling division with the symbols [x], and I will use <= for less than or equal to.



3 <= [X/50]



2 < X/50



100 < X



101 <= X



This is how my discrete structures professor requested how I complete this problem.






pigeonhole-principle




pigeonhole-principle






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 0:25







Nolan D.

















asked Dec 1 '18 at 1:27









Nolan D.Nolan D.

83




83








  • 2




    $begingroup$
    The answer is correct. Note that 101 doesn’t seem very reasonable considering that there are only 50 items to begin with.
    $endgroup$
    – platty
    Dec 1 '18 at 1:28










  • $begingroup$
    Why not tell everybody your thoughts about this question? A single answer is not helpful to you. People would be glad to check your reasoning and tell you which part is wrong. Please add your thoughts in the post, not in the comment section.
    $endgroup$
    – xbh
    Dec 1 '18 at 1:36












  • $begingroup$
    It's $2 < frac{X}{5}$ and $3 leq leftlceil frac{X}{5} rightrceil$.
    $endgroup$
    – Fabio Somenzi
    Dec 1 '18 at 1:51












  • $begingroup$
    Please edit the title of the question into something more descriptive.
    $endgroup$
    – Shaun
    Dec 1 '18 at 2:06










  • $begingroup$
    I think you read the question wrong: there are 10 items for each color, i.e. 10 blue, 10 red, 10 green, 10 yellow, and 10 purple.
    $endgroup$
    – xbh
    Dec 1 '18 at 2:56














  • 2




    $begingroup$
    The answer is correct. Note that 101 doesn’t seem very reasonable considering that there are only 50 items to begin with.
    $endgroup$
    – platty
    Dec 1 '18 at 1:28










  • $begingroup$
    Why not tell everybody your thoughts about this question? A single answer is not helpful to you. People would be glad to check your reasoning and tell you which part is wrong. Please add your thoughts in the post, not in the comment section.
    $endgroup$
    – xbh
    Dec 1 '18 at 1:36












  • $begingroup$
    It's $2 < frac{X}{5}$ and $3 leq leftlceil frac{X}{5} rightrceil$.
    $endgroup$
    – Fabio Somenzi
    Dec 1 '18 at 1:51












  • $begingroup$
    Please edit the title of the question into something more descriptive.
    $endgroup$
    – Shaun
    Dec 1 '18 at 2:06










  • $begingroup$
    I think you read the question wrong: there are 10 items for each color, i.e. 10 blue, 10 red, 10 green, 10 yellow, and 10 purple.
    $endgroup$
    – xbh
    Dec 1 '18 at 2:56








2




2




$begingroup$
The answer is correct. Note that 101 doesn’t seem very reasonable considering that there are only 50 items to begin with.
$endgroup$
– platty
Dec 1 '18 at 1:28




$begingroup$
The answer is correct. Note that 101 doesn’t seem very reasonable considering that there are only 50 items to begin with.
$endgroup$
– platty
Dec 1 '18 at 1:28












$begingroup$
Why not tell everybody your thoughts about this question? A single answer is not helpful to you. People would be glad to check your reasoning and tell you which part is wrong. Please add your thoughts in the post, not in the comment section.
$endgroup$
– xbh
Dec 1 '18 at 1:36






$begingroup$
Why not tell everybody your thoughts about this question? A single answer is not helpful to you. People would be glad to check your reasoning and tell you which part is wrong. Please add your thoughts in the post, not in the comment section.
$endgroup$
– xbh
Dec 1 '18 at 1:36














$begingroup$
It's $2 < frac{X}{5}$ and $3 leq leftlceil frac{X}{5} rightrceil$.
$endgroup$
– Fabio Somenzi
Dec 1 '18 at 1:51






$begingroup$
It's $2 < frac{X}{5}$ and $3 leq leftlceil frac{X}{5} rightrceil$.
$endgroup$
– Fabio Somenzi
Dec 1 '18 at 1:51














$begingroup$
Please edit the title of the question into something more descriptive.
$endgroup$
– Shaun
Dec 1 '18 at 2:06




$begingroup$
Please edit the title of the question into something more descriptive.
$endgroup$
– Shaun
Dec 1 '18 at 2:06












$begingroup$
I think you read the question wrong: there are 10 items for each color, i.e. 10 blue, 10 red, 10 green, 10 yellow, and 10 purple.
$endgroup$
– xbh
Dec 1 '18 at 2:56




$begingroup$
I think you read the question wrong: there are 10 items for each color, i.e. 10 blue, 10 red, 10 green, 10 yellow, and 10 purple.
$endgroup$
– xbh
Dec 1 '18 at 2:56










1 Answer
1






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oldest

votes


















2












$begingroup$

Think of the bins, one for each color. Keep putting the marbles in the binary color. If we got two for each color, that would be 5 * 2 = 10 marbles. The next marble,regarldess of color, would make one of the bins contain 3 marbles. Yes by chance we could use fewer marbles, but 11 is the worst case scenario.



This is the essence of the pigeonhole principle. The bins are the pigeonholes and, we see how we can fill the pigeonholes without the condition holding. Then we add one more object to obtain the solution.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I am not sure this is a clear answer to how the 101 was calculated.
    $endgroup$
    – NoChance
    Dec 1 '18 at 4:35











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Think of the bins, one for each color. Keep putting the marbles in the binary color. If we got two for each color, that would be 5 * 2 = 10 marbles. The next marble,regarldess of color, would make one of the bins contain 3 marbles. Yes by chance we could use fewer marbles, but 11 is the worst case scenario.



This is the essence of the pigeonhole principle. The bins are the pigeonholes and, we see how we can fill the pigeonholes without the condition holding. Then we add one more object to obtain the solution.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I am not sure this is a clear answer to how the 101 was calculated.
    $endgroup$
    – NoChance
    Dec 1 '18 at 4:35
















2












$begingroup$

Think of the bins, one for each color. Keep putting the marbles in the binary color. If we got two for each color, that would be 5 * 2 = 10 marbles. The next marble,regarldess of color, would make one of the bins contain 3 marbles. Yes by chance we could use fewer marbles, but 11 is the worst case scenario.



This is the essence of the pigeonhole principle. The bins are the pigeonholes and, we see how we can fill the pigeonholes without the condition holding. Then we add one more object to obtain the solution.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I am not sure this is a clear answer to how the 101 was calculated.
    $endgroup$
    – NoChance
    Dec 1 '18 at 4:35














2












2








2





$begingroup$

Think of the bins, one for each color. Keep putting the marbles in the binary color. If we got two for each color, that would be 5 * 2 = 10 marbles. The next marble,regarldess of color, would make one of the bins contain 3 marbles. Yes by chance we could use fewer marbles, but 11 is the worst case scenario.



This is the essence of the pigeonhole principle. The bins are the pigeonholes and, we see how we can fill the pigeonholes without the condition holding. Then we add one more object to obtain the solution.






share|cite|improve this answer









$endgroup$



Think of the bins, one for each color. Keep putting the marbles in the binary color. If we got two for each color, that would be 5 * 2 = 10 marbles. The next marble,regarldess of color, would make one of the bins contain 3 marbles. Yes by chance we could use fewer marbles, but 11 is the worst case scenario.



This is the essence of the pigeonhole principle. The bins are the pigeonholes and, we see how we can fill the pigeonholes without the condition holding. Then we add one more object to obtain the solution.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 1 '18 at 1:38









Joel PereiraJoel Pereira

73519




73519












  • $begingroup$
    I am not sure this is a clear answer to how the 101 was calculated.
    $endgroup$
    – NoChance
    Dec 1 '18 at 4:35


















  • $begingroup$
    I am not sure this is a clear answer to how the 101 was calculated.
    $endgroup$
    – NoChance
    Dec 1 '18 at 4:35
















$begingroup$
I am not sure this is a clear answer to how the 101 was calculated.
$endgroup$
– NoChance
Dec 1 '18 at 4:35




$begingroup$
I am not sure this is a clear answer to how the 101 was calculated.
$endgroup$
– NoChance
Dec 1 '18 at 4:35


















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