Combinatorics - pick every object in a set Suppose I have a set of N objects and I pick from this set (with replacement) n>N times. How many permutations are there that include all N objects?
I've tried a number of different things, but am kind of going in circles and expect it should be a pretty simple solution if you do it right. 
-- EDIT: My confusion of combinations vs. permutations has opened up a slightly bigger (or smaller perhaps) bag of worms on my methodology. The greater problem I am considering is as follows: Suppose you have N objects from which you draw with replacement. How many objects must you draw, n, s.t. the probability of having all N objects in your set is >p%, for some p. I thought I would determine the number of total permutations, N^n, and the set of all permutations that contain all N objects, say P(N). Then I would let q(n) = P(N)/N^n and find n s.t. q(n) >= p.
This started as a thought experiment and is much trickier than I expected. 
 A: Assuming the answer in my comments above comes back positive, that we are interested simply in the number of times each object is selected and not the order in which it was selected, suppose there are $N$ objects and you want select $n$ times with replacement, tallying how many times each was selected.  Let the $N$ objects be labeled $a_1, a_2, \dots a_N$ and let $x_i$ be the number of times $a_i$ was selected.
Note, $x_1 + x_2 + \dots + x_N = n$ and $x_i\geq 1$ for every $i$ since we selected objects a total of $n$ times and we are curious about the situation that we selected each object at least once.  The question is then to find the number of integral solutions to this.
To continue, consider a change of variable, $y_i = x_i - 1$.  Then $y_i \geq 0$ and $y_1 + y_2 + \dots + y_N = n - N$
By formula, http://en.wikipedia.org/wiki/Stars_and_bars_%28combinatorics%29
The number of combinations is then $\binom{N+(n-N)-1}{(n-N)-1} = \binom{ n-1}{n-N-1}$

In the case that it you are curious of the permutations instead of combinations, i.e., the order the elements were chosen in matters instead of simply the number of times each element was selected, you could take each possible combination (found above) and count the number of rearrangements of each and sum them all.  At each step, the number of permutations would be $\frac{n!}{x_1!x_2!\cdots x_N!}$.  With a computer you might be able to complete this train of thought but this doesn't seem to simplify well, so I'll scratch this method.
Instead thinking of an inclusion-exclusion approach, consider all sequences of length $n$ with entries chosen from $N$ elements $a_i$.  The generalized inclusion-exclusion principle essentially states (#none violate condition) = (#all possible no restriction) - (#at least one violation) + (#at least two violations) - (#at least three violations) ... +(#at least 2k violations) - (#at least 2k+1 violations) + ...
There are $N^n$ possible such sequences without restriction.  So our running total begins as $N^n$.
We subtract the number of sequences which at least don't include $a_1$ and subtract the number of sequences that at least don't include $a_2$ and $\dots$.
For each $a_i$ there are $(N-1)^n$ violating cases, and there are $N$ choices for which $a_i$ it was that was missing.  So, our running total is currently $N^n - N\cdot (N-1)^n$
Adding now the number of sequences which at least don't include $a_i$ and $a_j$ for $i\neq j$, there are $(N-2)^n$ violating cases, and there are $\binom{N}{2}$ number of ways to choose $i$ and $j$.  So, our running total is currently $N^n - N\cdot(N-1)^n + \binom{N}{2} (N-2)^n$
Continuing on to the general term, if there are $2k$ terms missing from the sequence, there are $(N-2k)^n$ number of sequences and $\binom{N}{2k}$ number of $2k$-tuples that might have been missing.
So, the answer would be:
$$N^n - N(N-1)^n + \binom{N}{2}(N-2)^n -\cdots + \binom{N}{2k}(N-2k)^n - \binom{N}{2k+1}(N-(2k+1))^n + \cdots \pm \binom{N}{N}(N-N)^n$$
Simplified into a single sum is $\sum\limits_{i=0}^N (-1)^i\binom{N}{i} (N-i)^n$
I personally do not recognize this as being able to be simplified further, but perhaps you might recognize it or someone else might later.
