How many combinations of three numbers using 1, 2, 3, and 4 exist? If you count (4 4 3) as one combination, you cannot count (4 3 4) as another.
My approach is $\dfrac{4^3}{3!}$, but obviously this does not work. I don't know why it doesn't work, and I don't know how I should solve the problem.
 A: Case 1: Three distinct numbers.
In this case there are four choices for the first number, three for the second, and two for the third, but each of these $24$ possibilites can be reached in $6$ different orders, so we must divide by $3!$, to get
$$\frac{4 \cdot 3 \cdot 2}{3!} = 4$$
(alternatively, there are simply $4$ ways to choose the number you leave out)
Case 2: Two of the same number, and a third different number.
There are four choices for the first number, and three for the second, for a total of
$$4\cdot 3 = 12$$
Case 3: All of the same number.
There are four choices for this number. 

The total is therefore
$$4 + 12 + 4 = \boxed{20}$$
A: Dividing by $3!$ means you only count $(1, 2, 3)$ one time, not once for each of the six permutations of $1,2,3$. That is good.
Unfortunately, dividing by $3!$ also means you count only $1/6$
of a combination for $(1,1,1)$.
You may do better if you count triples with all three digits different,
then the combinations with one pair and one digit different,
then the three-of-a-kind combinations.
