Combinatorics: Seating Problem

[Master1022]

Homework Statement

There are 7 people going to eat lunch at a table with 10 seats, arranged in two rows of 5 (2 x 5 arrangement). 5 people are wearing a red shirt, and the other 2 are wearing a blue shirt. If everyone sits down randomly, what is the probability that the number of people wearing a blue shirt sit across from one another?

Relevant Equations

Combinatorics

Hi,

I found this problem online and I wanted to see whether my solution was going along the correct lines or not?

Question: There are 7 people going to eat lunch at a table with 10 seats, arranged in two rows of 5 (2 x 5 arrangement). 5 people are wearing a red shirt, and the other 2 are wearing a blue shirt. If everyone sits down randomly, what is the probability that the number of people wearing a blue shirt sit across from one another?

Attempt:
I wanted to use a counting argument. Also, I think we assume that the people are indistinguishable from one another.

The total number of ways of them sitting down is $\begin{pmatrix} 10 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 8 \\ 5 \end{pmatrix}$.

Then the number of ways to satisfy the blue shirts sitting across from one another:
$$ = \text{number of locations for blue to sit across from one another} \times \text{number of ways for other 5 red people to sit in 8 remaining seats} $$
$$ = 5 \cdot \begin{pmatrix} 8 \\ 5 \end{pmatrix} $$

Thus the expression becomes:
$$ p = \frac{5 \cdot \begin{pmatrix} 8 \\ 5 \end{pmatrix}}{\begin{pmatrix} 10 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 8 \\ 5 \end{pmatrix}} = \frac{5}{\begin{pmatrix} 10 \\ 2 \end{pmatrix}} = \frac{1}{9} $$

Does this seem correct? Many thanks

 

[PeroK]

Imagine the first blue-shirt sits down. There is one place opposite him/her. The second blue-shirt sits at random and has a 1/9 chance of choosing the seat opposite.

 

[kuruman]

Imagine the first blue-shirt sits down. There is one place opposite him/her. The second blue-shirt sits at random and has a 1/9 chance of choosing the seat opposite.

The problem states that the people are seated randomly which means that red and blue shirts take seats in no particular sequence. I imagine a single file of non-interacting, randomly mixed red and blue shirts taking seats one at a time. Your solution assumes that the two blue shirts are at the front of the line. DId I miss something?

 

[pbuk]

DId I miss something?

Yes, you missed that the order sequence of seating is irrelevant because everyone sits down randomly.

 

[PeroK]

The problem states that the people are seated randomly which means that red and blue shirts take seats in no particular sequence. I imagine a single file of non-interacting, randomly mixed red and blue shirts taking seats one at a time. Your solution assumes that the two blue shirts are at the front of the line. DId I miss something?

The sequence with which people take their seats has no effect on the overall probability of sitting opposite someone else. In each case, every permutation of seating positions is equally likely. For example, let's compare the first and last people to sit down. The first has a choice of 10 seats, which they choose randomly, so they end up in each seat with equal probability. The last has a choice of only 4 seats, but each seat has an equal probability of remaining, so they also end up in each seat with equal probability.

It often simplifies these problems to think of the people you are interested in going first.

 

[kuruman]

It often simplifies these problems to think of the people you are interested in going first.

Thanks for the nicely detailed explanation. Thinking of the person of interest first confused me and I didn't consider that all microstates are equally likely.