- #1
Dragonfall
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- 4
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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