Drawing without replacement, order matters - probability function I'm not a mathematician, so the description of my problem might seem a little verbose:
In a multiple-choice test there is a set of three questions and a set of five possible answers. To each of the three questions one of the five answers is the correct one. No answer can be assigned twice. So three of the $5$ answers will be assigned to one of the three questions and two answers will be left.
I would now like to calculate the average score that would result from random answers if each correct answer yields one point.
If I'm not mistaken, in this case I have $n!(/n-r)!=60$ possible answer combinations, but I don't know what formula I use to calculate the probability of randomly hitting a combination with $0, 1, 2$ or $3$ correct answers. As far as I can see, this is a draw without replacement where the order matters, so I can neither treat it like a dice problem (e.g. "How likely are exactly three sixes in ten throws?" - binomial distribution) nor like a control sample ("How likely is it that exactly $5$ of these $100$ screws are defective?" - hypergeometric distribution), because then either the non-replacement or the order is not taken into account.
I wrote down all $60$ possible permutations and added up how many points each of them would yield. If I counted correctly, there ...
-... is one possible permutation that yields all three points,
-... are $6$ permutations that yield 2 points,
-... are $21$ permutations that yield 1 point,
-... are $32$ permutations that yield 0 points.
The table is in German, but I think it's not hard to guess what means what:
https://i.imgur.com/B669JzF.jpg
Can someone help me with that? Thanks in advance!
 A: To summarize the discussion in the comments.  The OP has, correctly, done the full enumeration of cases and from that we can deduce that the expectation is $E=.6$
Of course, that method is badly error prone and, in any case, would not be practical in situations wherein the numbers were much larger.
Linearity gives us a much more efficient way to proceed.  Indeed, let $\{X_i\}_{i=1}^3$ denote the scores attached to your three guesses in order.  Then of course $E[X_i]=\frac 15$ as any guess has a $\frac 15$ chance of getting a point (and a $\frac 45$ chance of getting $0$).  Then linearity tells us that $$E=E[X_1+X_2+X_3]=\sum_{i=1}^3 E[X_i]=\frac 35=.6$$
as confirmed by the full enumerator.
To stress: There is no claim that the $X_i$ are independent (indeed, they are not) but that is irrelevant, expectation is still linear.
A: 
I would now like to calculate the average score that would result from random answers if each correct answer yields one point.

You have $5 \times 4 \times 3 = 60$ possible responses to the test as a whole.
Let us assume your counting is right and there is $1$ combination giving three points, $6$ combinations yielding $2$ points, $21$ giving $1$ point and $32$ giving $0$ points.
Let $X$ be the variable denoting the points obtained from randomly answering the questions in the test. Then
$$\begin{cases}P(X=0) = \frac{32}{60}  \\ P(X=1) = 
\frac{21}{60} \\ P(X=2) =\frac{6}{60} \\ 
P(X=3) = \frac{1}{60}\end{cases}$$
which is equivalent to the discrete probability function
$$f(x) = \begin{cases}\frac{32}{60}  & x = 0\\ 
\frac{21}{60} & x = 1\\ \frac{6}{60} & x = 2\\ 
\frac{1}{60} & x =3\end{cases}$$
Then, by definition of expected value, we have
$$\mathbb{E}(X) = 0\times\frac{32}{60} + \frac{21}{60} + 2 \frac{6}{60} + 3 \times \frac{1}{60} = 0.6$$
So the expected value of $X$ is $0.6$.
