Permutation and Combination

[Smriti]

Can someone please help me out with these 4 questions

1)If n people are seated in a random manner in a row of n theatre saets, what is the prob. That two particular people A and B will be seated next to each other?


2)If k people are seated in  a random manner in a row, containg n seats, (n>k), what is the probability that the people will occupy k adjacent seats in a row ?

3)If k people are seated in a random manner in a circle, containing n chairs, (n>k) what is the probability that the people will occupy k adjacent chairs in the circle?


4)If n ppl are seated in a random manner containg 2n seats, what is the probability that no two people will occupy  adjacent seats?

[Chinni18]

Hi Smriti
Q.1
To count the number if ways to seat n people such that 2 of them are always together, count them as one person. Then the number of arrangements are (n-1)! Also there are 2 ways to arrange the the two people AB and BA. So total ways is 2(n-1)!
The total number of ways to arrange n people is n!
Thus the required probability is 2(n-1)! / n! =  2/n

[Chinni18]

Q.2
Total ways to seat k people in n places = C(n,k)*(k!)
To have them sitting adjacent to each other, you consider them as one unit and arrange this "unit" and the remaining (n-k) seats in (n-k+1)! ways. Within this unit they can be seated in k! ways
Thus probability = (n-k+1)! / C(n,k)

[Chinni18]

Q.3
First select k seats for these people in C(n,k) ways.
Once you have selected you can seat them in k! ways in the seats.
Thus, total number of ways of seating = C(n,k)*k!
Now to have them adjacent, consider them as 1 unit.
You can arrange the remaining seats [i.e n-k] and this "unit" in (n-k+1-1)! = (n-k)! ways
This is because the formula for circular arrangement of n items is (n-1)!

So your required probability is
(n-k)! / C(n,k)

[Chocolate Frog]

There are (n+1) ways of seating n people on 2n chairs such that no two people sit together. Of course, these n people can be arranged in n! ways among themselves. Hence, the required probability is (n+1)!/2nPn.