Probability/ permutation question.?

This is another thing I studied in my discrete math class last semester.

A re-arrangement of the elements of set so that no element ends up in the same location is called a "derangement".

It turns out that the number of possible derangements of a set with n elements is given by:

Dn = n! [1 - 1/1! + 1/2! - 1/3! + ... + (-1)^n 1/n!]

So, since the number of possible arrangements of n objects is n!, the probability of no two objects ending up in its original position is Dn/n!:

Pn = 1 - 1/1! + 1/2! - 1/3! + ... + (-1)^n 1/n!

And the probability that at least one item ends up in the same place is 1 - Pn:

1 - Pn = 1/1! - 1/2! + 1/3! + ... - (-1)^n 1/n!

An interesting fact is that the series representation for Pn is also the Taylor series expansion for e^(-1). So the more objects there are, the closer the probability of having no object in its original position gets to e^(-1), which is approximately 0.368.

Therefore, the probability that at least one item will end up in the same place approaches 1 - e^(-1) = 0.632, or 63.2%.

Well, I would say it is the probability that the "1" stays in the first position, plus the probability that "2" stays in the second position but "1" does not, plus, etc.

Out of "n" items, the chance that 1 stays put is 1/N, I'm pretty sure, because if you fill the first position, there is a 1/N chance that the "1" will be selected.

That means there is a (1-N)/N chance that 1 is NOT in the first position. So the probability that "2" stays put and "1" does not is 1/N*(1-N)/N

So our sum looks like

1/N + 1/N*(1-N)/N + etc. but I'm not sure how to turn this into a formula. I'll keep thinking and edit this but in case this much is helpful, I'll post.

inform you what, repost your tries to do those issues, then i will help. I recommend each and every physique else waits besides. Steve EDIT - inspite of the shown fact that i assume you took those solutions from the lower back of the e book or something, i will provide you the solutions besides. 8! / (6!2!) = 28 20 (4 a million-digit numbers, 4*4=sixteen 2-digit numbers) sure, 11!/ (2!2!)

P(non-match) = (n-1)!/n! = 1/n

P(match)= 1- P(non-match)= (n-1)/n

aw shoot Jim beat me to it