Re: Placing Balls in Urns and Expected values

There are n^n possibilities how to place n indexed balls into n indexed
urns, if balls are indexed, too.
From this number, there are n possibilities how to place all balls into
only one urn.
We have n!/(n-2)!1!1! how to chose two urn to place odd number of balls
in them, and eventually n!/(n-2)!2! , if the number of balls in both
urns is the same. These counts do not count with indexing of balls and
leave (n-2) urns empty.
If only urns are indexed, the sum of all such coefficients for all
partitions reduces to
(2m -1)!/m!(m-1).
The sum of both coefficients divided by this coefficient gives the
probability of (n-2) urns to be empty.
kunzmilan

.

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