Number of ways of distributing 6 objects to 6 persons
up vote
1
down vote
favorite
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 at 19:52
Jack Moody
16611
asked Nov 21 at 17:05
user3767495
1518
add a comment |
up vote
1
down vote
favorite
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 at 19:52
Jack Moody
16611
asked Nov 21 at 17:05
user3767495
1518
3
Where does the "$7$" in $7^6$ comes from?
– mlc
Nov 21 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 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 at 19:39
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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 at 19:52
Jack Moody
16611
asked Nov 21 at 17:05
user3767495
1518
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 at 19:52
Jack Moody
16611
asked Nov 21 at 17:05
user3767495
1518
edited Nov 21 at 19:52
Jack Moody
16611
asked Nov 21 at 17:05
user3767495
1518
edited Nov 21 at 19:52
Jack Moody
16611
edited Nov 21 at 19:52
Jack Moody
16611
edited Nov 21 at 19:52
Jack Moody
16611
16611
asked Nov 21 at 17:05
user3767495
1518
asked Nov 21 at 17:05
user3767495
1518
asked Nov 21 at 17:05
user3767495
1518
1518
3
Where does the "$7$" in $7^6$ comes from?
– mlc
Nov 21 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 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 at 19:39
add a comment |
3
Where does the "$7$" in $7^6$ comes from?
– mlc
Nov 21 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 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 at 19:39
3
3
Where does the "$7$" in $7^6$ comes from?
– mlc
Nov 21 at 19:37
Where does the "$7$" in $7^6$ comes from?
– mlc
Nov 21 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 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 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 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 at 19:39
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
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 at 19:48
idea
1
Please correct your first sentence.
– N. F. Taussig
Nov 21 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 at 3:33
@Idea-What Am I overcounting here with $7^6$ term?
– user3767495
Nov 22 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 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 at 4:45
|
show 13 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
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 at 19:48
idea
1
Please correct your first sentence.
– N. F. Taussig
Nov 21 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 at 3:33
@Idea-What Am I overcounting here with $7^6$ term?
– user3767495
Nov 22 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 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 at 4:45
|
show 13 more comments
up vote
1
down vote
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 at 19:48
idea
1
Please correct your first sentence.
– N. F. Taussig
Nov 21 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 at 3:33
@Idea-What Am I overcounting here with $7^6$ term?
– user3767495
Nov 22 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 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 at 4:45
|
show 13 more comments
up vote
1
down vote
up vote
1
down vote
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 at 19:48
idea
1
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 at 19:48
idea
1
edited Nov 22 at 4:25
answered Nov 21 at 19:48
idea
1
answered Nov 21 at 19:48
idea
1
answered Nov 21 at 19:48
idea
1
1
Please correct your first sentence.
– N. F. Taussig
Nov 21 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 at 3:33
@Idea-What Am I overcounting here with $7^6$ term?
– user3767495
Nov 22 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 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 at 4:45
|
show 13 more comments
Please correct your first sentence.
– N. F. Taussig
Nov 21 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 at 3:33
@Idea-What Am I overcounting here with $7^6$ term?
– user3767495
Nov 22 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 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 at 4:45
Please correct your first sentence.
– N. F. Taussig
Nov 21 at 20:24
Please correct your first sentence.
– N. F. Taussig
Nov 21 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 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 at 3:33
@Idea-What Am I overcounting here with $7^6$ term?
– user3767495
Nov 22 at 3:35
@Idea-What Am I overcounting here with $7^6$ term?
– user3767495
Nov 22 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 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 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 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 at 4:45
|
show 13 more comments
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 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 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 at 19:39