The concept of derangement is what you need to get this probabilty. Read Derangement - Wikipedia, the free encyclopedia.
suppose that a large number of people standing in a queue are asked to reorder themselves so that they line up in order of height. what is the approximate probability that at least one person is in the same position after the reordering as they were before?
i did up lists for up to 4 people and the results are:
1 person, Pr = 1 = 1
2 people, Pr = 1/2 = 0.5
3 people, Pr = 2/3 = 0.667
4 people, Pr = 5/8 = 0.625
i cannot for the life of me see a pattern, considering the 2 people probability means that the probabilities aren't even in decreasing order. i have done that particular probability correct, right?
also, the question has a hint:
ex = S xr
WHAT DOES THIS HAVE TO DO WITH THE QUESTION?!?!
ok. so the formula for number of derangements is
C:\Documents and Settings\Gabby\Desktop\derangement formula.JPG
that means that the approximate probability that at least one person is in the same position after the reordering as they were before is
1-C:\Documents and Settings\Gabby\Desktop\derangement formula.JPG
ok. i THINK i understand. its asking for the probability, so with the derangement formula, you dont need the n! at the start becuase you arent looking for a number, just a probability. then, the hint means that as r goes from 0 to infinite, the sum of (x^r)/r! equals e^x. so that means that if x = -1 such as in the derangement formula, as r goes from 0 to infinite, the sum of (-1)^r/r! actually equals e^-1 which equals 1/e. YAYY!!! thanks so much for your help!!