Probability of having exactly one empty container What's the probability that if we distribute N letters
among N drawers and presume that every single
configuration is equally likely (saturation is at
least 0 and at most N) then a single drawer
will be found empty?
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My progress so far:
Total number of cases: $N^{N}$
Reasoning:
It's the implication of the saturation condition.
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The number of favorable configurations:
$2 \cdot \binom{N}{2}\cdot (N - 2)!$
Reasoning:
I can choose $2$ drawers out of $N$ drawers in $\binom{N}{2}$ ways.
These two drawers serve as the empty one and the one containing two letters.
The factor $2$ is to consider the relative location of the two drawers.
$(N - 2)!$ is to take the permutation of the remaining drawers into account.
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I feel like something is mistaken.
Question: What's the route of correctness?
 A: Favorable Configurations:
$ \displaystyle 2 \cdot \binom{N}{2}\cdot \binom{N}{2} \cdot(N - 2)!$
We choose two drawers, one of them will hold two letters and one will be empty. Remaining $(N-2)$ drawers will hold one letter each. Now we choose two letters that will go into the drawer selected to keep two letters. Letters in remaining drawers can be arranged in $(N-2)!$ ways.
A: Your $N^N$ looks reasonable for the total number of equally likely arrangements.
One approach to the second part is to say that we choose $2$ of the $N$ items ${N\choose 2}$ ways and a drawer to put them in $N$ ways and another drawer to be empty $N-1$ ways and then organise the remaining $N-2$ drawers and $N-2$ items $(N-2)!$ ways, giving $${N\choose 2}N(N-1)\,(N-2)! = \frac{N(N-1)}2N!$$ possibilities, which we then divide by $N^N$ to get a probability.

*

*For $N=2$ this gives $\frac12$

*For $N=3$ this gives $\frac23$

*For $N=4$ this gives $\frac9{16}$

*For $N=5$ this gives $\frac{48}{125}$
which seems plausible
