# Thread: Permutations &Combinations in Probability

1. ## Permutations &Combinations in Probability

8 people of different heights are to be seated in a row. what is the probability that (a) there are at least 3 people sitting between the tallest and shortest? (b) the tallest and shortest are sitting next to each other?

Hi, if you could please explain how to do the problem above I would really appreciate it! Stats is my weak point in math and I would really appreciate if you could explain the logic behind the work you did, if not its okay! Thanks anyway!

2. Total ways to seat these 8 people: 8!

a) Let's count how many ways we can seat our tallest and shortest with at least 3 people between them. I'm going to have the tallest to the left and shortest to the right. We'll double our count because I could have an equal number of representations in reverse.

3 space configs: TxxxSxxx, xTxxxSxx, xxTxxxSx, xxxTxxxS
Each config has 6! ways to seat the remaining people; 4*6! ways total.

4 space configs: TxxxxSxx, xTxxxxSx, xxTxxxxS
Each config has 6! ways to seat the remaining people; 3*6! ways total.

5 space configs: TxxxxxSx, xTxxxxxS
Each config has 6! ways to seat the remaining people; 2*6! ways total.

6 space configs: TxxxxxxS
This config has 6! ways to seat the remaining people; 6! ways total.

It appears we have 10*6! ways to do this, remember to double. So we've got 20*6! ways to handle this.

Probability: (20*6!)/(8!) = 20/56 = 5/14

b) I think this one is much easier to count. Use a similar method to what I've done above and good luck!