Probability of getting a word after scrambling it I rearrange the letters in machinery. What's the probability that I end up with machinery?
Is the answer just $1/9!$ since there's only one way in which I can get this combination? And nine letters means $9!$ orderings?
 A: Yes, you are correct. One nitpick: You say combination, but what you really mean here is permutation. A combination is a selection of items where we don’t care about the order, and a permutation is a selection where we do care. You also used the word ‘orderings’ which essentially means the same thing as permutations, so you’re correct there.
For example, $machinery$ and $mcineyahr$ are the same combination, but they are different permutations.
Although in this case it’s clear what we mean, it isn’t always—so it’s worth being careful.
Two ways of thinking about this:
1) Compute the probability at each stage of construction. For the first letter, we have a $\frac{1}{9}$ chance of being correct since there are $9$ letters left and exactly one of them is the correct one. When we place the next letter, with $8$ letters left, the chance is $\frac{1}{8}$ of this letter being correct. Continue on like this until all of the letters are placed. Since we are seeking the probability of all of these events occurring in this order, we multiply the probabilities to get
$$\frac{1}{9} \cdot \frac{1}{8} \cdot \dots \cdot \frac{1}{2} \cdot \frac{1}{1} = \frac{1}{9!}$$
2) Consider the probability of getting the correct permutation to be a fraction of the total number of possible permutations. The total number of possible permutations here is $9!$, and there is one one correct permutation, so we have $\frac{1}{9!}$ as you say.
These two methods are essentially equivalent, but in some cases one will be easier to compute than the other.
(As noted in the comments: if the set has any duplicate items, in this case if the word has duplicate letters, then there will be a different number of possible permutations)
