Mixing
Number of ways of distributing 6 objects to 6 persons
What is the number of ways of distributing 6 objects to 6 persons such that at least one of them does not get anything.
I worked like
Without restriction total ways= $7^6$
Number of cases in which all persons get something=$6!$
Number of cases where at least one person does not get anything=$7^6-6!$
Am I correct?
Also is the probability that at least one person doesn’t get any object equal to $frac{7^6-6!}{7^6}$?
combinatorics
edited Nov 21 '18 at 19:52
Jack Moody
16611
asked Nov 21 '18 at 17:05
user3767495
3448
add a comment |
What is the number of ways of distributing 6 objects to 6 persons such that at least one of them does not get anything.
I worked like
Without restriction total ways= $7^6$
Number of cases in which all persons get something=$6!$
Number of cases where at least one person does not get anything=$7^6-6!$
Am I correct?
Also is the probability that at least one person doesn’t get any object equal to $frac{7^6-6!}{7^6}$?
combinatorics
edited Nov 21 '18 at 19:52
Jack Moody
16611
asked Nov 21 '18 at 17:05
user3767495
3448
3
Where does the "$7$" in $7^6$ comes from?
– mlc
Nov 21 '18 at 19:37
Without restriction, I believe it would be $6^6$ since you have 6 people to give each object (assuming objects and people are unique).
– Jack Moody
Nov 21 '18 at 19:38
i think OP means $7^{th}$ way is when a person does not get any object, which is incorrect.
– idea
Nov 21 '18 at 19:39
add a comment |
What is the number of ways of distributing 6 objects to 6 persons such that at least one of them does not get anything.
I worked like
Without restriction total ways= $7^6$
Number of cases in which all persons get something=$6!$
Number of cases where at least one person does not get anything=$7^6-6!$
Am I correct?
Also is the probability that at least one person doesn’t get any object equal to $frac{7^6-6!}{7^6}$?
combinatorics
edited Nov 21 '18 at 19:52
Jack Moody
16611
asked Nov 21 '18 at 17:05
user3767495
3448
What is the number of ways of distributing 6 objects to 6 persons such that at least one of them does not get anything.
I worked like
Without restriction total ways= $7^6$
Number of cases in which all persons get something=$6!$
Number of cases where at least one person does not get anything=$7^6-6!$
Am I correct?
Also is the probability that at least one person doesn’t get any object equal to $frac{7^6-6!}{7^6}$?
combinatorics
combinatorics
edited Nov 21 '18 at 19:52
Jack Moody
16611
asked Nov 21 '18 at 17:05
user3767495
3448
edited Nov 21 '18 at 19:52
Jack Moody
16611
asked Nov 21 '18 at 17:05
user3767495
3448
edited Nov 21 '18 at 19:52
Jack Moody
16611
edited Nov 21 '18 at 19:52
Jack Moody
16611
edited Nov 21 '18 at 19:52
Jack Moody
16611
16611
asked Nov 21 '18 at 17:05
user3767495
3448
asked Nov 21 '18 at 17:05
user3767495
3448
asked Nov 21 '18 at 17:05
user3767495
3448
3448
3
Where does the "$7$" in $7^6$ comes from?
– mlc
Nov 21 '18 at 19:37
Without restriction, I believe it would be $6^6$ since you have 6 people to give each object (assuming objects and people are unique).
– Jack Moody
Nov 21 '18 at 19:38
i think OP means $7^{th}$ way is when a person does not get any object, which is incorrect.
– idea
Nov 21 '18 at 19:39
add a comment |
3
Where does the "$7$" in $7^6$ comes from?
– mlc
Nov 21 '18 at 19:37
Without restriction, I believe it would be $6^6$ since you have 6 people to give each object (assuming objects and people are unique).
– Jack Moody
Nov 21 '18 at 19:38
i think OP means $7^{th}$ way is when a person does not get any object, which is incorrect.
– idea
Nov 21 '18 at 19:39
3
3
Where does the "$7$" in $7^6$ comes from?
– mlc
Nov 21 '18 at 19:37
Where does the "$7$" in $7^6$ comes from?
– mlc
Nov 21 '18 at 19:37
Without restriction, I believe it would be $6^6$ since you have 6 people to give each object (assuming objects and people are unique).
– Jack Moody
Nov 21 '18 at 19:38
Without restriction, I believe it would be $6^6$ since you have 6 people to give each object (assuming objects and people are unique).
– Jack Moody
Nov 21 '18 at 19:38
i think OP means $7^{th}$ way is when a person does not get any object, which is incorrect.
– idea
Nov 21 '18 at 19:39
i think OP means $7^{th}$ way is when a person does not get any object, which is incorrect.
– idea
Nov 21 '18 at 19:39
add a comment |
1 Answer
1
active
oldest
votes
Total number of ways to distribute $6$ objects among $6$ people such that each person gets exactly one object is $6!$
Proof: A particular object has $6$ choices, i.e. $6$ people to select $1$ from. Similarly, every object has $6$ choices, so total ways are $6^6$.
You might have selected a person and calculated choices for him: $6$ for $6$ objects and $1$ when he does not get any object. This gave you $7^6$, which is incorrect.
So, total ways (without restriction) are $6^6$.
And, number of ways when each person gets $1$ object is $6!$, which you got.
So, required number of ways are $$6^6-6!$$ giving you a probability of $$frac{6^6-6!}{6^6}$$
answered Nov 21 '18 at 19:48
idea
2,15441025
Please correct your first sentence.
– N. F. Taussig
Nov 21 '18 at 20:24
@idea-I thought like for each object, I have 6 choices to give to 6 persons and 1 more choice to give nothing to any of 6 persons. So, it was like 7 choices for each object. Imagine like 6 persons are 6 boxes, and there is 1 another box called "No-one" box Now, each object has 7 choices to go into a box. So, from here I calculated ways to distribute 6 objects to 6 persons without any restriction.
– user3767495
Nov 22 '18 at 3:33
@Idea-What Am I overcounting here with $7^6$ term?
– user3767495
Nov 22 '18 at 3:35
1
There are $6^6$ ways to assign persons to the objects when the $6$ objects may each receive one from $6$ persons with repetitions allowed. Among these ways, some persons may not be assigned to any objects because several objects have been assigned the same person. @user3767495
– Graham Kemp
Nov 22 '18 at 4:28
2
$7^6$, counts ways six people can make a choice to take one from seven objects (with the seventh choice a "no" object). This may leave you with several people taking the same object, and some objects not taken, but no person will ever have more than one object. @user3767495
– Graham Kemp
Nov 22 '18 at 4:45
|
show 13 more comments
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%2f3008026%2fnumber-of-ways-of-distributing-6-objects-to-6-persons%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Total number of ways to distribute $6$ objects among $6$ people such that each person gets exactly one object is $6!$
Proof: A particular object has $6$ choices, i.e. $6$ people to select $1$ from. Similarly, every object has $6$ choices, so total ways are $6^6$.
You might have selected a person and calculated choices for him: $6$ for $6$ objects and $1$ when he does not get any object. This gave you $7^6$, which is incorrect.
So, total ways (without restriction) are $6^6$.
And, number of ways when each person gets $1$ object is $6!$, which you got.
So, required number of ways are $$6^6-6!$$ giving you a probability of $$frac{6^6-6!}{6^6}$$
answered Nov 21 '18 at 19:48
idea
2,15441025
Please correct your first sentence.
– N. F. Taussig
Nov 21 '18 at 20:24
@idea-I thought like for each object, I have 6 choices to give to 6 persons and 1 more choice to give nothing to any of 6 persons. So, it was like 7 choices for each object. Imagine like 6 persons are 6 boxes, and there is 1 another box called "No-one" box Now, each object has 7 choices to go into a box. So, from here I calculated ways to distribute 6 objects to 6 persons without any restriction.
– user3767495
Nov 22 '18 at 3:33
@Idea-What Am I overcounting here with $7^6$ term?
– user3767495
Nov 22 '18 at 3:35
1
There are $6^6$ ways to assign persons to the objects when the $6$ objects may each receive one from $6$ persons with repetitions allowed. Among these ways, some persons may not be assigned to any objects because several objects have been assigned the same person. @user3767495
– Graham Kemp
Nov 22 '18 at 4:28
2
$7^6$, counts ways six people can make a choice to take one from seven objects (with the seventh choice a "no" object). This may leave you with several people taking the same object, and some objects not taken, but no person will ever have more than one object. @user3767495
– Graham Kemp
Nov 22 '18 at 4:45
|
show 13 more comments
Total number of ways to distribute $6$ objects among $6$ people such that each person gets exactly one object is $6!$
Proof: A particular object has $6$ choices, i.e. $6$ people to select $1$ from. Similarly, every object has $6$ choices, so total ways are $6^6$.
You might have selected a person and calculated choices for him: $6$ for $6$ objects and $1$ when he does not get any object. This gave you $7^6$, which is incorrect.
So, total ways (without restriction) are $6^6$.
And, number of ways when each person gets $1$ object is $6!$, which you got.
So, required number of ways are $$6^6-6!$$ giving you a probability of $$frac{6^6-6!}{6^6}$$
answered Nov 21 '18 at 19:48
idea
2,15441025
Please correct your first sentence.
– N. F. Taussig
Nov 21 '18 at 20:24
@idea-I thought like for each object, I have 6 choices to give to 6 persons and 1 more choice to give nothing to any of 6 persons. So, it was like 7 choices for each object. Imagine like 6 persons are 6 boxes, and there is 1 another box called "No-one" box Now, each object has 7 choices to go into a box. So, from here I calculated ways to distribute 6 objects to 6 persons without any restriction.
– user3767495
Nov 22 '18 at 3:33
@Idea-What Am I overcounting here with $7^6$ term?
– user3767495
Nov 22 '18 at 3:35
1
There are $6^6$ ways to assign persons to the objects when the $6$ objects may each receive one from $6$ persons with repetitions allowed. Among these ways, some persons may not be assigned to any objects because several objects have been assigned the same person. @user3767495
– Graham Kemp
Nov 22 '18 at 4:28
2
$7^6$, counts ways six people can make a choice to take one from seven objects (with the seventh choice a "no" object). This may leave you with several people taking the same object, and some objects not taken, but no person will ever have more than one object. @user3767495
– Graham Kemp
Nov 22 '18 at 4:45
|
show 13 more comments
Total number of ways to distribute $6$ objects among $6$ people such that each person gets exactly one object is $6!$
Proof: A particular object has $6$ choices, i.e. $6$ people to select $1$ from. Similarly, every object has $6$ choices, so total ways are $6^6$.
You might have selected a person and calculated choices for him: $6$ for $6$ objects and $1$ when he does not get any object. This gave you $7^6$, which is incorrect.
So, total ways (without restriction) are $6^6$.
And, number of ways when each person gets $1$ object is $6!$, which you got.
So, required number of ways are $$6^6-6!$$ giving you a probability of $$frac{6^6-6!}{6^6}$$
answered Nov 21 '18 at 19:48
idea
2,15441025
Total number of ways to distribute $6$ objects among $6$ people such that each person gets exactly one object is $6!$
Proof: A particular object has $6$ choices, i.e. $6$ people to select $1$ from. Similarly, every object has $6$ choices, so total ways are $6^6$.
You might have selected a person and calculated choices for him: $6$ for $6$ objects and $1$ when he does not get any object. This gave you $7^6$, which is incorrect.
So, total ways (without restriction) are $6^6$.
And, number of ways when each person gets $1$ object is $6!$, which you got.
So, required number of ways are $$6^6-6!$$ giving you a probability of $$frac{6^6-6!}{6^6}$$
answered Nov 21 '18 at 19:48
idea
2,15441025
edited Nov 22 '18 at 4:25
answered Nov 21 '18 at 19:48
idea
2,15441025
answered Nov 21 '18 at 19:48
idea
2,15441025
answered Nov 21 '18 at 19:48
idea
2,15441025
2,15441025
Please correct your first sentence.
– N. F. Taussig
Nov 21 '18 at 20:24
@idea-I thought like for each object, I have 6 choices to give to 6 persons and 1 more choice to give nothing to any of 6 persons. So, it was like 7 choices for each object. Imagine like 6 persons are 6 boxes, and there is 1 another box called "No-one" box Now, each object has 7 choices to go into a box. So, from here I calculated ways to distribute 6 objects to 6 persons without any restriction.
– user3767495
Nov 22 '18 at 3:33
@Idea-What Am I overcounting here with $7^6$ term?
– user3767495
Nov 22 '18 at 3:35
1
There are $6^6$ ways to assign persons to the objects when the $6$ objects may each receive one from $6$ persons with repetitions allowed. Among these ways, some persons may not be assigned to any objects because several objects have been assigned the same person. @user3767495
– Graham Kemp
Nov 22 '18 at 4:28
2
$7^6$, counts ways six people can make a choice to take one from seven objects (with the seventh choice a "no" object). This may leave you with several people taking the same object, and some objects not taken, but no person will ever have more than one object. @user3767495
– Graham Kemp
Nov 22 '18 at 4:45
|
show 13 more comments
Please correct your first sentence.
– N. F. Taussig
Nov 21 '18 at 20:24
@idea-I thought like for each object, I have 6 choices to give to 6 persons and 1 more choice to give nothing to any of 6 persons. So, it was like 7 choices for each object. Imagine like 6 persons are 6 boxes, and there is 1 another box called "No-one" box Now, each object has 7 choices to go into a box. So, from here I calculated ways to distribute 6 objects to 6 persons without any restriction.
– user3767495
Nov 22 '18 at 3:33
@Idea-What Am I overcounting here with $7^6$ term?
– user3767495
Nov 22 '18 at 3:35
1
There are $6^6$ ways to assign persons to the objects when the $6$ objects may each receive one from $6$ persons with repetitions allowed. Among these ways, some persons may not be assigned to any objects because several objects have been assigned the same person. @user3767495
– Graham Kemp
Nov 22 '18 at 4:28
2
$7^6$, counts ways six people can make a choice to take one from seven objects (with the seventh choice a "no" object). This may leave you with several people taking the same object, and some objects not taken, but no person will ever have more than one object. @user3767495
– Graham Kemp
Nov 22 '18 at 4:45
Please correct your first sentence.
– N. F. Taussig
Nov 21 '18 at 20:24
Please correct your first sentence.
– N. F. Taussig
Nov 21 '18 at 20:24
@idea-I thought like for each object, I have 6 choices to give to 6 persons and 1 more choice to give nothing to any of 6 persons. So, it was like 7 choices for each object. Imagine like 6 persons are 6 boxes, and there is 1 another box called "No-one" box Now, each object has 7 choices to go into a box. So, from here I calculated ways to distribute 6 objects to 6 persons without any restriction.
– user3767495
Nov 22 '18 at 3:33
@idea-I thought like for each object, I have 6 choices to give to 6 persons and 1 more choice to give nothing to any of 6 persons. So, it was like 7 choices for each object. Imagine like 6 persons are 6 boxes, and there is 1 another box called "No-one" box Now, each object has 7 choices to go into a box. So, from here I calculated ways to distribute 6 objects to 6 persons without any restriction.
– user3767495
Nov 22 '18 at 3:33
@Idea-What Am I overcounting here with $7^6$ term?
– user3767495
Nov 22 '18 at 3:35
@Idea-What Am I overcounting here with $7^6$ term?
– user3767495
Nov 22 '18 at 3:35
1
1
There are $6^6$ ways to assign persons to the objects when the $6$ objects may each receive one from $6$ persons with repetitions allowed. Among these ways, some persons may not be assigned to any objects because several objects have been assigned the same person. @user3767495
– Graham Kemp
Nov 22 '18 at 4:28
There are $6^6$ ways to assign persons to the objects when the $6$ objects may each receive one from $6$ persons with repetitions allowed. Among these ways, some persons may not be assigned to any objects because several objects have been assigned the same person. @user3767495
– Graham Kemp
Nov 22 '18 at 4:28
2
2
$7^6$, counts ways six people can make a choice to take one from seven objects (with the seventh choice a "no" object). This may leave you with several people taking the same object, and some objects not taken, but no person will ever have more than one object. @user3767495
– Graham Kemp
Nov 22 '18 at 4:45
$7^6$, counts ways six people can make a choice to take one from seven objects (with the seventh choice a "no" object). This may leave you with several people taking the same object, and some objects not taken, but no person will ever have more than one object. @user3767495
– Graham Kemp
Nov 22 '18 at 4:45
|
show 13 more comments
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3008026%2fnumber-of-ways-of-distributing-6-objects-to-6-persons%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
3
Where does the "$7$" in $7^6$ comes from?
– mlc
Nov 21 '18 at 19:37
Without restriction, I believe it would be $6^6$ since you have 6 people to give each object (assuming objects and people are unique).
– Jack Moody
Nov 21 '18 at 19:38
i think OP means $7^{th}$ way is when a person does not get any object, which is incorrect.
– idea
Nov 21 '18 at 19:39