Allen McIntosh wrote:
>
> (Followup directed to sci.stat.edu)
>
> In article <[EMAIL PROTECTED]>,
> DeLa <[EMAIL PROTECTED]> wrote:
> >Maybe it is a bit off topic, but I couldn't find a ng that better matches my
> >issue. Which is...
> >
> >We are a group of nine people. We want each of us to buy a present for one
> >other member. We have a hat with names and each one picks a name. What is
> >the chance of at least one picking his or her own name?
Allen McIntosh wrote:
> Let A be the set of permutations in which the first object gets mapped
> to itself, B be the set in which the second object gets mapped to itself,
> and so on. Write out a simple expression for the set of permutations
> in which no object gets mapped to itself. Then write out an expression
> for the number of elements in the set. (I'm assuming you are taking
> a course with some elementary probability content, and the you've
> seen things like #(A union B) = #(A) + #(B) - #(AB).)
This last hint is *very* important. I think it would be fair to add:
..and thngs like #(A U B U C) = #A + #B + #C - $AB -$BC -#CA + #ABC
and to point out that this is still a difficult problem!
-Robert Dawson
.
.
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