Counting Principles and Probability

[ScPpL]Shree

1. The problem statement, all variables and given/known data
Either People of different heights are to be seated in a row. The shortest and tallest in this group are not seated at either end. What is the probability that:

a) the tallest and shortest persons are sitting next to each other? (Ans: 1/3)
b) there is one person sitting between the tallest and shortest? (Ans: 4/15)

2. Relevant equations

combinations, permutations, factorials

3. The attempt at a solution

i didnt even b cuz i dont get a. here is my solution for part a:

_ _ _ _ _ _ _ _ are the seats. the tallest and shortest cant be at the ends, so that leaves 6

_ _ _ _ _ _. the number of ways the tallest and shortest can sit together is:

(4!)(2) for the tallest/shortest person sitting in the middle 4 seats, and the other one is going to be on the seat that is left or right of the first 1. This added with 2 for the ends of this section of 6.

the number of ways the the above scenario can happen is: 6! (for the tallest/shortest) X 5! (for the other one) X 6! (for the rest of the people)

therefore the total probability for part a is: [(4!)(2) + 2]6! divided by 6!5!6! = 1/864, but thats wrong :/
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution

Karlx

Hi Shree.
You must calculate carefully the number of ways the eight people can be seated, with this two conditions:
1) People will be seated in a row, that is, there are two ends.
2) The shortest and the tallest can't be seated in the ends of the row.

Try to seat first of all the shortest and the tallest. In how many ways can they be seated ?
After that, let sit the other people. In how many ways can they do it ?

To calculate the number of ways the eight people can be seated in order to satisfy a), you may follow the same way, sitting first the shortest and the tallest and then the other people.

I hope this could help you.