Number of outcomes times probability of failure is number of failures? I was hoping someone could spread some insight on why this is true?  Suppose that in some situation, there are n! number of possible outcomes.  Also, suppose that p is the probability of failure of all n events (i.e. a matching problem where no one picks their correct match), then it results that the number of outcomes that are failures is n!*p.
I was maybe trying to relate it to the expected value for a binomial random variable, but perhaps we cannot relate expected value to here, since n!*p is the number of ways, not the expected number of ways?
Thanks!

If we have a sample space of $N$ equally likely outcomes, and the number of outcomes that result in failure in $F$, then the probability $p$ of failure is given by 
$$p=\frac{F}{N}.$$
This can be rewritten as $F=Np$. 
Remark: The example you have in mind is that we consider the $n!$ permutations of the set $\{1,2,3,\dots,n\}$. Under the right assumptions, these $n!$ permutations are equally likely. Let $D_n$ be the number of derangements of the set $\{1,2,\dots,n\}$. If $p$ is the probability that a randomly chosen permutation is a derangement, then $p=\frac{D_n}{n!}$ and therefore $D_n=n!p$.  
