Getting a "straight" in dice rolls Suppose that you have $k$ dice, each with $N$ sides, where $k\geq N$.  The definition of a straight is when all $k$ dice are rolled, there is at least one die revealing each number from $1$ to $N$.  
Given the pair $(k,N)$, what is the probability that any particular roll will give a straight?
 A: I'm assuming that you're familiar with the inclusion-exclusion principle. In this particular problem, you need to determine the probability of a straight, or in other words, the complement of the event that atleast $1$ number does not appear in the throws. Let that probability be $P$. Also, let $p(j)$ be the probability that you're excluding atleast $j$ numbers from the the throw. Then 
$$p(1)={N \choose 1}\left(\frac{N-1}{N}\right)^k$$
$$p(2)={N \choose 2}\left(\frac{N-2}{N}\right)^k$$
and so on.
Then $$P=1-p(1)+p(2)+ \cdots+(-1)^np(n)$$
Which is exactly what you wanted.
A: An equivalent problem is fitting $N-1$ bars between $k$ stars so that there is one star between each bar.  Here the stars represent the $k$ dice, and the bars are the borders of boxes which represent the $N$ values the dice can take.
There are ${k-1\choose N-1}$ ways to do this.
There are ${N+k-1\choose N-1}$ ways to arrange the stars and bars without this condition.
So the probability of getting a straight is: $$\mathcal{\Large P}(\text{Straight})=\frac{k-1\choose N-1}{N+k-1\choose N-1}=\frac{(k-1)!k!}{(k-N)!(N+k-1)!}, \quad\forall k\geq N$$
A: What about simply $\dfrac{^k\text{P}_n(k-n)!}{n^k}$, since $k \ge n$. If $k=n$ then we simply get $\dfrac{n!}{n^k}$?
Since there are only $n$ sides to the dice and if $k=n$, we are asking the total number of permutations of $n$ different values which is $^n\text{P}_n=n!$ divided by all the possible outcomes of $n^k$. If $k>n$ we just replace $^n\text{P}_n$ with $^k\text{P}_n$ assume some values besides $n$ different possible sides are included thus multiplying the total possible outcomes with at least a straight.
