y**t
发帖数: 50 | 1 Consider permutations on 6 letters.It's the
probability that a permutation without fixed points.
You can calcualte it inductively.assume it's f(n)
f(2)=1,f(3)=2,f(4)=9,f(5)=44,f(6)=265
so I think it's 265/6!=53/144 | d****b
发帖数: 24 | 2 fix the envelope.
Then for the letters, total permutations are 6!.
at least one is correct c(6,1)*5!
at lease two is correct c(6,2)*4!
at least three is correct c(6,3)*3!
at lease four is correct c(6,4)*2!
at least five is correct c(6,5)*1!
all six is correct c(6,6)*0!
By Bayes' formula
Therefore, the solution is
{c(6,0)*6!-c(6,1)*5!+c(6,2)*4!-c(6,3)*3!+c(6,4)*2!-c(6,5)*1!+c(6,6)*0!}/6!
={720-720+360-120+30-6+1}/720
=265/720 |
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