Alternative way to write combinatory probabilities Suppose we have a bag of $20$ balls, each ball is uniquely numbered from $1$ to $20$.
Say that we choose any $5$ unique numbers at random between $1$ and $20$ inclusive, before any ball has been drawn from the bag.
$5$ Balls are then drawn without replacement.
I know the probability of all $5$ being correctly guessed can be expressed as:
$$5/20 \times 4/19 \times 3/18 \times 2/17 \times 1/16$$
I also know the probability of correctly guessing $4$ balls out of the $5$ is:
$$\frac{C(5, 4) \times C(20-5, 1)}{C(20, 5)},$$
which is the number of combinations of $4$ numbers in every $5$ ball draw, over the total.
What I'm looking for in particular: is there a way to express $P(4$ balls guessed out of $5)$ without using combinations as I have done just above this?
Every topic I've found on the subject only shows answers using combinations.
 A: A mild variant of your "all right" calculation will do it. Imagine drawing the balls one at a time. Write Y if the ball you get matches one of your numbers, and N if it doesn't. Then we can get exactly $4$ right in $5$ ways: (i) NYYYY, (ii) YNYYY, (iii) YYNYY, (iv) YYYNY, or (v) YYYYN.
Let us calculate  the probability of NYYYY. The probability the first is a N is $\frac{15}{20}$. Given that the first was a N, the probability the second is a Y is $\frac{5}{19}$. Given the first was a N and the second was a Y, the probability the third is a Y is $\frac{4}{18}$. and so on. So the probability of NYYYY is
$$\frac{15}{20}\cdot\frac{5}{19}\cdot\frac{4}{18}\cdot\frac{3}{17}\cdot\frac{2}{16}.$$
Now we find the probability of YNYYY. There is an easy way to do it (all sequences of $5$ balls are equally likely, so the answer will be the same as for NYYYY). Or there is the hard way, computing just as for NYYYY. If we do that, we get
$$\frac{5}{20}\cdot\frac{15}{19}\cdot\frac{4}{18}\cdot\frac{3}{17}\cdot\frac{2}{16}.$$ 
This is the same number as for NYYYY.
The other $3$ cases each give the same probability. So the probability of $4$ right and $1$ wrong is
$$(5)\left(\frac{15}{20}\cdot\frac{5}{19}\cdot\frac{4}{18}\cdot\frac{3}{17}\cdot\frac{2}{16}\right).$$
