# Probability homework problem

1. Apr 22, 2012

### th4450

1. The problem statement, all variables and given/known data
Bobo, Coco, and 8 other students are arranged to sit in 2 rows of 5 students. If these 10 students take their seat randomly, find the probability that Bobo and Coco are sitting in the same row but not next to each other.

3. The attempt at a solution
Bobo and Coco together have 12 ways to sit, they 2 can exchange, other 8 students sit randomly.
Probability = $\frac{12×2×8!}{10!}$ = $\frac{4}{15}$

Is this correct? Thank you!!

2. Apr 22, 2012

### HallsofIvy

Staff Emeritus
Re: Probability

Correct answer to what question? Where did you include the "sitting in the same row"? Or "not next to each other?

3. Apr 23, 2012

### th4450

Re: Probability

I mean, say the seats are arranged like this:
1 2 3 4 5
6 7 8 9 10

Bobo and Coco sitting in the same row but not next to each other, they can take:
1,3
2,4
3,5
1,4
2,5
1,5
6,8
7,9
8,10
6,9
7,10
6,10
totally 12 ways.

The above arrangement can be for Bobo|Coco or Coco|Bobo, so times 2.

Is my answer correct? Thanks again.

4. Apr 23, 2012

### HallsofIvy

Staff Emeritus
Re: Probability

Yes, that is correct- here is how I would have done it- a different way to get the same answer:
Coco can sit anywhere in a given row. Then Bobo must sit in the same row but not next to Coco.
There is a slight complication here- If Coco is sitting in an end seat there are 3 seats where Bobo can sit. But if Coco is sitting in any other seat, there are only 2 seats where Coco can sit. That is the number possible ways for both Bobo and Coco to sit are 2*3+ 3*2= 12. Since there are 2 rows there are 2*12= 24 ways Bobo and Coco can sit in the two rows.