- #1

Dragonfall

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

I think it should be

[tex]\binom{n}{i}\frac{(n-i-1)!}{n!}[/tex]

Is it right?

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

[tex]\sum_{i=0}^{n}i\binom{n}{i}\frac{(n-i-1)!}{n!}[/tex]

But it doesn't sum to 1, I think

I think it should be

[tex]\binom{n}{i}\frac{(n-i-1)!}{n!}[/tex]

Is it right?

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

[tex]\sum_{i=0}^{n}i\binom{n}{i}\frac{(n-i-1)!}{n!}[/tex]

But it doesn't sum to 1, I think

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