What Are the Probabilities of Different Seating Arrangements at a Party?

[stunner5000pt]

Homework Statement

you and your two friends are with 5 other people who are randomly assigned to two tables of four.
a) calculate the number of possible seating arrangements (where actual placement of your chair does not matter, only relevant placement of individuals at a table)
b) compute the probability of you sit next to your two friends
c) compute the probability that you sit at a different table as your two friends
d) compute the probability that you sit next to your two friends when you know you have been assigned to the same table

2. The attempt at a solution
a) The total number of ways is 8!

b) Confused because if this is a round table (doesnt specify that) has a situation where you don't sit directly next to your friends, as one may be a person away.
Assuming that that is a relevant issue here:
You x 2 possibly ways of friends seated x 5! x 8 (8 different spots for you to sit in)

c) when friends are at a different table: your friends cannot be on your table which is not 1 (you) x 5 x 4 x 3 x (other table) 4 x3x2x1 x 8 (spots for you to sit in)

d) if you know you will sit on the same table, there are two ways that your friends would sit next to you and 4 other ways that they sit on the same table and one of them is NOT next to you so 2/6 or 1/3

Is this correct?

Thank you

 

[Ray Vickson]

Homework Statement

you and your two friends are with 5 other people who are randomly assigned to two tables of four.
a) calculate the number of possible seating arrangements (where actual placement of your chair does not matter, only relevant placement of individuals at a table)
b) compute the probability of you sit next to your two friends
c) compute the probability that you sit at a different table as your two friends
d) compute the probability that you sit next to your two friends when you know you have been assigned to the same table

2. The attempt at a solution
a) The total number of ways is 8!

b) Confused because if this is a round table (doesnt specify that) has a situation where you don't sit directly next to your friends, as one may be a person away.
Assuming that that is a relevant issue here:
You x 2 possibly ways of friends seated x 5! x 8 (8 different spots for you to sit in)

c) when friends are at a different table: your friends cannot be on your table which is not 1 (you) x 5 x 4 x 3 x (other table) 4 x3x2x1 x 8 (spots for you to sit in)

d) if you know you will sit on the same table, there are two ways that your friends would sit next to you and 4 other ways that they sit on the same table and one of them is NOT next to you so 2/6 or 1/3

Is this correct?

Thank you

For (b): Imagine you first sit somewhere. What is the probability that one of your friends will sit on your immediate right? Then what is the probability that your other friend will sit on your immediate left?

For (c): after you sit (first), what is the probability your two friends sit at a different table from you? You can compute this from the hypergeometric distribution, where the remaining 7 people fall into two classes: I = friend (2 members) and II = non-friend (5 members). You want the probability that in a sample of size 3 drawn from the 7 remaining people, no members of class I occur.

 

[haruspex]

Not sure I understand what (a) is asking for. It might mean that the orientation of each table does not matter. I.e. having decided who sits at which table, it's only the relative positions at each table that matter.
b and c ask for probabilities, but you only quote combinations; I assume you're aware of that. So I'll interpret your answers to these as being in relation to 8! total combinations.
I agree with your answers to b, c and d.