Number of ways of distributing 6 objects to 6 persons
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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
1688
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
1688
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
1688
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
1688
edited Nov 21 at 19:52
Jack Moody
16611
asked Nov 21 at 17:05
user3767495
1688
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
1688
asked Nov 21 at 17:05
user3767495
1688
asked Nov 21 at 17:05
user3767495
1688
1688
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
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oldest
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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
1,96231024
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
Your Answer
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1 Answer
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1 Answer
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oldest
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oldest
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up vote
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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
1,96231024
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
1,96231024
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
1,96231024
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
1,96231024
edited Nov 22 at 4:25
answered Nov 21 at 19:48
idea
1,96231024
answered Nov 21 at 19:48
idea
1,96231024
answered Nov 21 at 19:48
idea
1,96231024
1,96231024
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
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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