n persons have alloted seats while they take seats at random.What is the probability that

i) exactly k persons have found the alloted seats;

ii) at most k persons have found the alloted seats;

iii) at least p persons have found the alloted seats .

1)If n no. of people sit randomly, they can sit in n! ways.
Out of these k no. of people sit in their alloted seats.
This means that (n-k) no. of people sit randomly.
No. of ways in which they sit in places other than the ones alloted to them

=(n-k-1)(n-k-2)....(1)=(n-k-1)!
Thus,
required probability=(n-k-1)! / n!

PLEAZZZZZ RATE IF CORRECT....

No aysh!!!!!!!

Actually I have the answer of the first part and it is (n-k-1)! / (2k⁴) but I don't know how it has come.I agree with you and I think some other cases in the problem have to be considered as well.Well please try!!!!!!!!!!!!!!!!!!!! for the ii) and iii).