# Probablity of fixed points in permutations

1. Feb 14, 2008

### Dragonfall

Randomly permute (1,...,n). What is the probability that exactly i points are fixed?

I think it should be

$$\binom{n}{i}\frac{(n-i-1)!}{n!}$$

Is it right?

If so, is the expected number of fixed points (I know it's 1):

$$\sum_{i=0}^{n}i\binom{n}{i}\frac{(n-i-1)!}{n!}$$

But it doesn't sum to 1, I think

Last edited: Feb 14, 2008
2. Feb 14, 2008

### EnumaElish

How is (n-i-1)! defined when i = n?

3. Feb 14, 2008

### Dragonfall

It isn't. Nevermind. I was approaching the problem in a unnecessarily hard way.