probability of groupings of people around a table
assuming X people sit down around a (round) table, Y people have black shirts and X-Y people have white shirts, what is the probability that the two clusters of shirts are grouped together (i.e., all the black shirts sit together, all white shirts together) around the table?
i've thought about this using standard permutations with circular probability (i.e., (X p Y) / (X-1)!). i've also thought about it sequentially (i.e., someone with a black shirt sits down, so looking at the probability that another black shirt sits down, and another, etc). i get similar results with both methods but do not believe either approach is correct.
thanks in advance for the help and sorry to bore with rudimentary math.
probability combinatorics permutations
asked Dec 11 '18 at 4:45
Will Cumberland
11
add a comment |
assuming X people sit down around a (round) table, Y people have black shirts and X-Y people have white shirts, what is the probability that the two clusters of shirts are grouped together (i.e., all the black shirts sit together, all white shirts together) around the table?
i've thought about this using standard permutations with circular probability (i.e., (X p Y) / (X-1)!). i've also thought about it sequentially (i.e., someone with a black shirt sits down, so looking at the probability that another black shirt sits down, and another, etc). i get similar results with both methods but do not believe either approach is correct.
thanks in advance for the help and sorry to bore with rudimentary math.
probability combinatorics permutations
asked Dec 11 '18 at 4:45
Will Cumberland
11
Welcome to MathSE. When you pose a question here, it is expected that you include your own thoughts on the problem. Please edit the question to tell us what you know, show what you have attempted, and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
Dec 11 '18 at 10:17
thanks, i revised the original question to include my two original approaches
– Will Cumberland
Dec 11 '18 at 17:58
add a comment |
assuming X people sit down around a (round) table, Y people have black shirts and X-Y people have white shirts, what is the probability that the two clusters of shirts are grouped together (i.e., all the black shirts sit together, all white shirts together) around the table?
i've thought about this using standard permutations with circular probability (i.e., (X p Y) / (X-1)!). i've also thought about it sequentially (i.e., someone with a black shirt sits down, so looking at the probability that another black shirt sits down, and another, etc). i get similar results with both methods but do not believe either approach is correct.
thanks in advance for the help and sorry to bore with rudimentary math.
probability combinatorics permutations
asked Dec 11 '18 at 4:45
Will Cumberland
11
assuming X people sit down around a (round) table, Y people have black shirts and X-Y people have white shirts, what is the probability that the two clusters of shirts are grouped together (i.e., all the black shirts sit together, all white shirts together) around the table?
i've thought about this using standard permutations with circular probability (i.e., (X p Y) / (X-1)!). i've also thought about it sequentially (i.e., someone with a black shirt sits down, so looking at the probability that another black shirt sits down, and another, etc). i get similar results with both methods but do not believe either approach is correct.
thanks in advance for the help and sorry to bore with rudimentary math.
probability combinatorics permutations
probability combinatorics permutations
asked Dec 11 '18 at 4:45
Will Cumberland
11
asked Dec 11 '18 at 4:45
Will Cumberland
11
edited Dec 11 '18 at 17:58
asked Dec 11 '18 at 4:45
Will Cumberland
11
asked Dec 11 '18 at 4:45
Will Cumberland
11
asked Dec 11 '18 at 4:45
Will Cumberland
11
11
Welcome to MathSE. When you pose a question here, it is expected that you include your own thoughts on the problem. Please edit the question to tell us what you know, show what you have attempted, and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
Dec 11 '18 at 10:17
thanks, i revised the original question to include my two original approaches
– Will Cumberland
Dec 11 '18 at 17:58
add a comment |
Welcome to MathSE. When you pose a question here, it is expected that you include your own thoughts on the problem. Please edit the question to tell us what you know, show what you have attempted, and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
Dec 11 '18 at 10:17
thanks, i revised the original question to include my two original approaches
– Will Cumberland
Dec 11 '18 at 17:58
Welcome to MathSE. When you pose a question here, it is expected that you include your own thoughts on the problem. Please edit the question to tell us what you know, show what you have attempted, and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
Dec 11 '18 at 10:17
Welcome to MathSE. When you pose a question here, it is expected that you include your own thoughts on the problem. Please edit the question to tell us what you know, show what you have attempted, and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
Dec 11 '18 at 10:17
thanks, i revised the original question to include my two original approaches
– Will Cumberland
Dec 11 '18 at 17:58
thanks, i revised the original question to include my two original approaches
– Will Cumberland
Dec 11 '18 at 17:58
add a comment |
1 Answer
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Form a group of all black shirts and add it to the the rest forming a total of $X-Y+1$ objects. Now each of the people inside the group can be shuffled around in $Y!$ . So total number of ways is
$$n=Y!cdot (X-Y)!$$
Now the total number of ways to seat $N$ people around a table are
$$N=(X-1)!$$
So, the probability is
$$P=frac{Y!cdot (X-Y)!}{(X-1)!}$$
This is only valid for the scenario where there is at least $1$ black t-shirt and at least $1$ white t-shirt. For the extreme cases i.e. $Y=X$ and $Y=0$, the probability is simply $1$ as there is no possibility of even forming $2$ groups.
answered Dec 11 '18 at 18:31
Sauhard Sharma
71714
Check your first sentence. What objects are you describing?
– N. F. Taussig
Dec 11 '18 at 20:21
Here I am considering each white t-shirt as an object and the collective black t-shirt as another object in the group
– Sauhard Sharma
Dec 12 '18 at 3:51
It might be clearer if you observe that you have two blocks of people, one in black shirts and one in white shirts, who can be placed around the table in $(2 - 1)! = 1$ distinguishable way, so we just have to worry about internal arrangements.
– N. F. Taussig
Dec 12 '18 at 9:39
I thought about that initially and in the end you arrive at the same thing, except you have to consider the similar possibility that one of the groups will be empty.
– Sauhard Sharma
Dec 12 '18 at 10:50
If a group is empty, then we are left with one group, and there is only one way to seat one object at a table. Similarly, if both groups are empty, there is only one way to no objects at a table. That said, I now see why you approached the problem the way you did.
– N. F. Taussig
Dec 12 '18 at 11:15
add a comment |
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1 Answer
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Form a group of all black shirts and add it to the the rest forming a total of $X-Y+1$ objects. Now each of the people inside the group can be shuffled around in $Y!$ . So total number of ways is
$$n=Y!cdot (X-Y)!$$
Now the total number of ways to seat $N$ people around a table are
$$N=(X-1)!$$
So, the probability is
$$P=frac{Y!cdot (X-Y)!}{(X-1)!}$$
This is only valid for the scenario where there is at least $1$ black t-shirt and at least $1$ white t-shirt. For the extreme cases i.e. $Y=X$ and $Y=0$, the probability is simply $1$ as there is no possibility of even forming $2$ groups.
answered Dec 11 '18 at 18:31
Sauhard Sharma
71714
Check your first sentence. What objects are you describing?
– N. F. Taussig
Dec 11 '18 at 20:21
Here I am considering each white t-shirt as an object and the collective black t-shirt as another object in the group
– Sauhard Sharma
Dec 12 '18 at 3:51
It might be clearer if you observe that you have two blocks of people, one in black shirts and one in white shirts, who can be placed around the table in $(2 - 1)! = 1$ distinguishable way, so we just have to worry about internal arrangements.
– N. F. Taussig
Dec 12 '18 at 9:39
I thought about that initially and in the end you arrive at the same thing, except you have to consider the similar possibility that one of the groups will be empty.
– Sauhard Sharma
Dec 12 '18 at 10:50
If a group is empty, then we are left with one group, and there is only one way to seat one object at a table. Similarly, if both groups are empty, there is only one way to no objects at a table. That said, I now see why you approached the problem the way you did.
– N. F. Taussig
Dec 12 '18 at 11:15
add a comment |
Form a group of all black shirts and add it to the the rest forming a total of $X-Y+1$ objects. Now each of the people inside the group can be shuffled around in $Y!$ . So total number of ways is
$$n=Y!cdot (X-Y)!$$
Now the total number of ways to seat $N$ people around a table are
$$N=(X-1)!$$
So, the probability is
$$P=frac{Y!cdot (X-Y)!}{(X-1)!}$$
This is only valid for the scenario where there is at least $1$ black t-shirt and at least $1$ white t-shirt. For the extreme cases i.e. $Y=X$ and $Y=0$, the probability is simply $1$ as there is no possibility of even forming $2$ groups.
answered Dec 11 '18 at 18:31
Sauhard Sharma
71714
Check your first sentence. What objects are you describing?
– N. F. Taussig
Dec 11 '18 at 20:21
Here I am considering each white t-shirt as an object and the collective black t-shirt as another object in the group
– Sauhard Sharma
Dec 12 '18 at 3:51
It might be clearer if you observe that you have two blocks of people, one in black shirts and one in white shirts, who can be placed around the table in $(2 - 1)! = 1$ distinguishable way, so we just have to worry about internal arrangements.
– N. F. Taussig
Dec 12 '18 at 9:39
I thought about that initially and in the end you arrive at the same thing, except you have to consider the similar possibility that one of the groups will be empty.
– Sauhard Sharma
Dec 12 '18 at 10:50
If a group is empty, then we are left with one group, and there is only one way to seat one object at a table. Similarly, if both groups are empty, there is only one way to no objects at a table. That said, I now see why you approached the problem the way you did.
– N. F. Taussig
Dec 12 '18 at 11:15
add a comment |
Form a group of all black shirts and add it to the the rest forming a total of $X-Y+1$ objects. Now each of the people inside the group can be shuffled around in $Y!$ . So total number of ways is
$$n=Y!cdot (X-Y)!$$
Now the total number of ways to seat $N$ people around a table are
$$N=(X-1)!$$
So, the probability is
$$P=frac{Y!cdot (X-Y)!}{(X-1)!}$$
This is only valid for the scenario where there is at least $1$ black t-shirt and at least $1$ white t-shirt. For the extreme cases i.e. $Y=X$ and $Y=0$, the probability is simply $1$ as there is no possibility of even forming $2$ groups.
answered Dec 11 '18 at 18:31
Sauhard Sharma
71714
Form a group of all black shirts and add it to the the rest forming a total of $X-Y+1$ objects. Now each of the people inside the group can be shuffled around in $Y!$ . So total number of ways is
$$n=Y!cdot (X-Y)!$$
Now the total number of ways to seat $N$ people around a table are
$$N=(X-1)!$$
So, the probability is
$$P=frac{Y!cdot (X-Y)!}{(X-1)!}$$
This is only valid for the scenario where there is at least $1$ black t-shirt and at least $1$ white t-shirt. For the extreme cases i.e. $Y=X$ and $Y=0$, the probability is simply $1$ as there is no possibility of even forming $2$ groups.
answered Dec 11 '18 at 18:31
Sauhard Sharma
71714
answered Dec 11 '18 at 18:31
Sauhard Sharma
71714
answered Dec 11 '18 at 18:31
Sauhard Sharma
71714
answered Dec 11 '18 at 18:31
Sauhard Sharma
71714
71714
Check your first sentence. What objects are you describing?
– N. F. Taussig
Dec 11 '18 at 20:21
Here I am considering each white t-shirt as an object and the collective black t-shirt as another object in the group
– Sauhard Sharma
Dec 12 '18 at 3:51
It might be clearer if you observe that you have two blocks of people, one in black shirts and one in white shirts, who can be placed around the table in $(2 - 1)! = 1$ distinguishable way, so we just have to worry about internal arrangements.
– N. F. Taussig
Dec 12 '18 at 9:39
I thought about that initially and in the end you arrive at the same thing, except you have to consider the similar possibility that one of the groups will be empty.
– Sauhard Sharma
Dec 12 '18 at 10:50
If a group is empty, then we are left with one group, and there is only one way to seat one object at a table. Similarly, if both groups are empty, there is only one way to no objects at a table. That said, I now see why you approached the problem the way you did.
– N. F. Taussig
Dec 12 '18 at 11:15
add a comment |
Check your first sentence. What objects are you describing?
– N. F. Taussig
Dec 11 '18 at 20:21
Here I am considering each white t-shirt as an object and the collective black t-shirt as another object in the group
– Sauhard Sharma
Dec 12 '18 at 3:51
It might be clearer if you observe that you have two blocks of people, one in black shirts and one in white shirts, who can be placed around the table in $(2 - 1)! = 1$ distinguishable way, so we just have to worry about internal arrangements.
– N. F. Taussig
Dec 12 '18 at 9:39
I thought about that initially and in the end you arrive at the same thing, except you have to consider the similar possibility that one of the groups will be empty.
– Sauhard Sharma
Dec 12 '18 at 10:50
If a group is empty, then we are left with one group, and there is only one way to seat one object at a table. Similarly, if both groups are empty, there is only one way to no objects at a table. That said, I now see why you approached the problem the way you did.
– N. F. Taussig
Dec 12 '18 at 11:15
Check your first sentence. What objects are you describing?
– N. F. Taussig
Dec 11 '18 at 20:21
Check your first sentence. What objects are you describing?
– N. F. Taussig
Dec 11 '18 at 20:21
Here I am considering each white t-shirt as an object and the collective black t-shirt as another object in the group
– Sauhard Sharma
Dec 12 '18 at 3:51
Here I am considering each white t-shirt as an object and the collective black t-shirt as another object in the group
– Sauhard Sharma
Dec 12 '18 at 3:51
It might be clearer if you observe that you have two blocks of people, one in black shirts and one in white shirts, who can be placed around the table in $(2 - 1)! = 1$ distinguishable way, so we just have to worry about internal arrangements.
– N. F. Taussig
Dec 12 '18 at 9:39
It might be clearer if you observe that you have two blocks of people, one in black shirts and one in white shirts, who can be placed around the table in $(2 - 1)! = 1$ distinguishable way, so we just have to worry about internal arrangements.
– N. F. Taussig
Dec 12 '18 at 9:39
I thought about that initially and in the end you arrive at the same thing, except you have to consider the similar possibility that one of the groups will be empty.
– Sauhard Sharma
Dec 12 '18 at 10:50
I thought about that initially and in the end you arrive at the same thing, except you have to consider the similar possibility that one of the groups will be empty.
– Sauhard Sharma
Dec 12 '18 at 10:50
If a group is empty, then we are left with one group, and there is only one way to seat one object at a table. Similarly, if both groups are empty, there is only one way to no objects at a table. That said, I now see why you approached the problem the way you did.
– N. F. Taussig
Dec 12 '18 at 11:15
If a group is empty, then we are left with one group, and there is only one way to seat one object at a table. Similarly, if both groups are empty, there is only one way to no objects at a table. That said, I now see why you approached the problem the way you did.
– N. F. Taussig
Dec 12 '18 at 11:15
add a comment |
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Welcome to MathSE. When you pose a question here, it is expected that you include your own thoughts on the problem. Please edit the question to tell us what you know, show what you have attempted, and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
Dec 11 '18 at 10:17
thanks, i revised the original question to include my two original approaches
– Will Cumberland
Dec 11 '18 at 17:58