How many people need to turn up to be sure of making at least one team?
$begingroup$
So pigeonhole principle states when there are m objects to be divided into n sets then at least one contains r+1 objects. i.e m > nr
In this question the m objects should be 12 months in 5 people so 12>5*3?
How do you arrive at 49?
combinatorics discrete-mathematics pigeonhole-principle elementary-probability
asked Nov 27 '18 at 10:39
PumpkinpeachPumpkinpeach
628
$endgroup$
add a comment |
$begingroup$
So pigeonhole principle states when there are m objects to be divided into n sets then at least one contains r+1 objects. i.e m > nr
In this question the m objects should be 12 months in 5 people so 12>5*3?
How do you arrive at 49?
combinatorics discrete-mathematics pigeonhole-principle elementary-probability
asked Nov 27 '18 at 10:39
PumpkinpeachPumpkinpeach
628
$endgroup$
add a comment |
$begingroup$
So pigeonhole principle states when there are m objects to be divided into n sets then at least one contains r+1 objects. i.e m > nr
In this question the m objects should be 12 months in 5 people so 12>5*3?
How do you arrive at 49?
combinatorics discrete-mathematics pigeonhole-principle elementary-probability
asked Nov 27 '18 at 10:39
PumpkinpeachPumpkinpeach
628
$endgroup$
So pigeonhole principle states when there are m objects to be divided into n sets then at least one contains r+1 objects. i.e m > nr
In this question the m objects should be 12 months in 5 people so 12>5*3?
How do you arrive at 49?
combinatorics discrete-mathematics pigeonhole-principle elementary-probability
combinatorics discrete-mathematics pigeonhole-principle elementary-probability
asked Nov 27 '18 at 10:39
PumpkinpeachPumpkinpeach
628
asked Nov 27 '18 at 10:39
PumpkinpeachPumpkinpeach
628
asked Nov 27 '18 at 10:39
PumpkinpeachPumpkinpeach
628
asked Nov 27 '18 at 10:39
PumpkinpeachPumpkinpeach
628
asked Nov 27 '18 at 10:39
PumpkinpeachPumpkinpeach
628
628
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The value we are trying to find in this problem is $m$, the number of people required to have at least one team of five people.
We want at least one team with $5$ players, so $r + 1 = 5 implies r = 4$.
The number of teams is the number of months, so $n = 12$.
Hence, according to your formula $m > 12 cdot 4 = 48$. The smallest such value is $m = 49$.
answered Nov 27 '18 at 10:47
N. F. TaussigN. F. Taussig
44k93356
$endgroup$
$begingroup$
How can number of teams be number of months?
$endgroup$
– Pumpkinpeach
Nov 27 '18 at 13:11
2
$begingroup$
It says that the members of each team must be born in the same month of the year. Hence, the number of possible teams is equal to the number of months.
$endgroup$
– N. F. Taussig
Nov 27 '18 at 13:17
add a comment |
$begingroup$
There are 12 possible months. Let's imagine "the worst" situation for finding team quickly: each new person has the most "unpopular" month of birth - so, ppl are distributed "uniformly" between month. How many ppl you need to have 4 of each type? 4*12=48. And then any new person (+1) will have 1 of 12 possible months of birth and 4+1=5 (team is made).
If ppl won't distribute uniformly, then you will need 48 or less (because if one group has 3, then another has 5). 4*12+1=49 is maximum, then.
answered Nov 27 '18 at 10:51
Kelly ShepphardKelly Shepphard
2298
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3015624%2fhow-many-people-need-to-turn-up-to-be-sure-of-making-at-least-one-team%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The value we are trying to find in this problem is $m$, the number of people required to have at least one team of five people.
We want at least one team with $5$ players, so $r + 1 = 5 implies r = 4$.
The number of teams is the number of months, so $n = 12$.
Hence, according to your formula $m > 12 cdot 4 = 48$. The smallest such value is $m = 49$.
answered Nov 27 '18 at 10:47
N. F. TaussigN. F. Taussig
44k93356
$endgroup$
$begingroup$
How can number of teams be number of months?
$endgroup$
– Pumpkinpeach
Nov 27 '18 at 13:11
2
$begingroup$
It says that the members of each team must be born in the same month of the year. Hence, the number of possible teams is equal to the number of months.
$endgroup$
– N. F. Taussig
Nov 27 '18 at 13:17
add a comment |
$begingroup$
The value we are trying to find in this problem is $m$, the number of people required to have at least one team of five people.
We want at least one team with $5$ players, so $r + 1 = 5 implies r = 4$.
The number of teams is the number of months, so $n = 12$.
Hence, according to your formula $m > 12 cdot 4 = 48$. The smallest such value is $m = 49$.
answered Nov 27 '18 at 10:47
N. F. TaussigN. F. Taussig
44k93356
$endgroup$
$begingroup$
How can number of teams be number of months?
$endgroup$
– Pumpkinpeach
Nov 27 '18 at 13:11
2
$begingroup$
It says that the members of each team must be born in the same month of the year. Hence, the number of possible teams is equal to the number of months.
$endgroup$
– N. F. Taussig
Nov 27 '18 at 13:17
add a comment |
$begingroup$
The value we are trying to find in this problem is $m$, the number of people required to have at least one team of five people.
We want at least one team with $5$ players, so $r + 1 = 5 implies r = 4$.
The number of teams is the number of months, so $n = 12$.
Hence, according to your formula $m > 12 cdot 4 = 48$. The smallest such value is $m = 49$.
answered Nov 27 '18 at 10:47
N. F. TaussigN. F. Taussig
44k93356
$endgroup$
The value we are trying to find in this problem is $m$, the number of people required to have at least one team of five people.
We want at least one team with $5$ players, so $r + 1 = 5 implies r = 4$.
The number of teams is the number of months, so $n = 12$.
Hence, according to your formula $m > 12 cdot 4 = 48$. The smallest such value is $m = 49$.
answered Nov 27 '18 at 10:47
N. F. TaussigN. F. Taussig
44k93356
answered Nov 27 '18 at 10:47
N. F. TaussigN. F. Taussig
44k93356
answered Nov 27 '18 at 10:47
N. F. TaussigN. F. Taussig
44k93356
answered Nov 27 '18 at 10:47
N. F. TaussigN. F. Taussig
44k93356
44k93356
$begingroup$
How can number of teams be number of months?
$endgroup$
– Pumpkinpeach
Nov 27 '18 at 13:11
2
$begingroup$
It says that the members of each team must be born in the same month of the year. Hence, the number of possible teams is equal to the number of months.
$endgroup$
– N. F. Taussig
Nov 27 '18 at 13:17
add a comment |
$begingroup$
How can number of teams be number of months?
$endgroup$
– Pumpkinpeach
Nov 27 '18 at 13:11
2
$begingroup$
It says that the members of each team must be born in the same month of the year. Hence, the number of possible teams is equal to the number of months.
$endgroup$
– N. F. Taussig
Nov 27 '18 at 13:17
$begingroup$
How can number of teams be number of months?
$endgroup$
– Pumpkinpeach
Nov 27 '18 at 13:11
$begingroup$
How can number of teams be number of months?
$endgroup$
– Pumpkinpeach
Nov 27 '18 at 13:11
2
2
$begingroup$
It says that the members of each team must be born in the same month of the year. Hence, the number of possible teams is equal to the number of months.
$endgroup$
– N. F. Taussig
Nov 27 '18 at 13:17
$begingroup$
It says that the members of each team must be born in the same month of the year. Hence, the number of possible teams is equal to the number of months.
$endgroup$
– N. F. Taussig
Nov 27 '18 at 13:17
add a comment |
$begingroup$
There are 12 possible months. Let's imagine "the worst" situation for finding team quickly: each new person has the most "unpopular" month of birth - so, ppl are distributed "uniformly" between month. How many ppl you need to have 4 of each type? 4*12=48. And then any new person (+1) will have 1 of 12 possible months of birth and 4+1=5 (team is made).
If ppl won't distribute uniformly, then you will need 48 or less (because if one group has 3, then another has 5). 4*12+1=49 is maximum, then.
answered Nov 27 '18 at 10:51
Kelly ShepphardKelly Shepphard
2298
$endgroup$
add a comment |
$begingroup$
There are 12 possible months. Let's imagine "the worst" situation for finding team quickly: each new person has the most "unpopular" month of birth - so, ppl are distributed "uniformly" between month. How many ppl you need to have 4 of each type? 4*12=48. And then any new person (+1) will have 1 of 12 possible months of birth and 4+1=5 (team is made).
If ppl won't distribute uniformly, then you will need 48 or less (because if one group has 3, then another has 5). 4*12+1=49 is maximum, then.
answered Nov 27 '18 at 10:51
Kelly ShepphardKelly Shepphard
2298
$endgroup$
add a comment |
$begingroup$
There are 12 possible months. Let's imagine "the worst" situation for finding team quickly: each new person has the most "unpopular" month of birth - so, ppl are distributed "uniformly" between month. How many ppl you need to have 4 of each type? 4*12=48. And then any new person (+1) will have 1 of 12 possible months of birth and 4+1=5 (team is made).
If ppl won't distribute uniformly, then you will need 48 or less (because if one group has 3, then another has 5). 4*12+1=49 is maximum, then.
answered Nov 27 '18 at 10:51
Kelly ShepphardKelly Shepphard
2298
$endgroup$
There are 12 possible months. Let's imagine "the worst" situation for finding team quickly: each new person has the most "unpopular" month of birth - so, ppl are distributed "uniformly" between month. How many ppl you need to have 4 of each type? 4*12=48. And then any new person (+1) will have 1 of 12 possible months of birth and 4+1=5 (team is made).
If ppl won't distribute uniformly, then you will need 48 or less (because if one group has 3, then another has 5). 4*12+1=49 is maximum, then.
answered Nov 27 '18 at 10:51
Kelly ShepphardKelly Shepphard
2298
answered Nov 27 '18 at 10:51
Kelly ShepphardKelly Shepphard
2298
answered Nov 27 '18 at 10:51
Kelly ShepphardKelly Shepphard
2298
answered Nov 27 '18 at 10:51
Kelly ShepphardKelly Shepphard
2298
2298
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3015624%2fhow-many-people-need-to-turn-up-to-be-sure-of-making-at-least-one-team%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
