How many valid permutations are there? What concept is this? Initially I thought it was a simple problem, but I can't figure out.
The password will have $1$, $2$, $3$, and a fourth digit. We can do casework on the value of the fourth digit:
Case 1: The fourth digit is not $1$, $2$, or $3$.
In this case, there are $7$ choices for the value of the fourth digit. For each of these $7$ choices, there are $4!$ ways to arrange the digits into a password:
$$7 \cdot 4! \,= \,168$$
Case 2: The fourth digit is $1$.
In this case, there would be $4!$ ways to arrange the digits if the $1$s were distinguishable, but since the $1$s are not distinguishable, we are overcounting by a factor of $2$ and we need to divide:
$$\frac{4!}{2} = 12$$
Case 3: The fourth digit is $2$.
In this case, the reasoning is the same as for Case 2:
$$\frac{4!}{2} = 12$$
Case 4: The fourth digit is $3$.
In this case, the reasoning is the same as for Cases 1 and 2:
$$\frac{4!}{2} = 12$$
This gives a final answer of
$$168 + 12 \cdot 3 \, = \, \boxed{204\,}$$
Let $A$ be the set of all passwords that contain 1, $B$, those contain 2 and $C$, those contain 3. We need the number of elements in $A\cap B \cap C$. Now, by inclusion exclusion formula, \begin{align*} |A^c \cup B^c \cup C^c| &= \binom{3}{1}9^4 - \binom{3}{2}8^4 + \binom{3}{3} 7^4 \end{align*} Hence \begin{align*} |A\cap B \cap C| = 10^4 - |A^c \cup B^c \cup C^c| = 204 \end{align*}