How many $4$ digit numbers can be created from digits: $0$, $3$, $4$, $6$, $7$, $9$? I got this as:
$$\dfrac{6!}{2!} = 360$$
To exclude numbers starting with zero: $$\dfrac {5!}{2!} = 60$$
Final result therefore: $$360 - 60 = 300$$
Is this correct?
There are some additional tasks I can't figure out:
a) In how many of them no digit is present more than once?
b) How many of them contain only one digit three times?
c) How many of them are even?
d) How many of them can be divided by $3$ with digit sum of $18$?
If these are digits on a four-digit combination lock and $0$ is good in any place, then you have $6^4$ possibilities. If the thousands digit needs to be more than $0$, then there are $5 \cdot 6^3$ possibilities.
But let's continue assuming that a leading zero is all right.
You got (a) already: $6 \cdot 5 \cdot 4 \cdot 3 = 360$.
For (b), pick the digit that occurs three times ($6$) and where they appear in the number ($_4C_3 = 4$). Then pick the other digit ($5$). So, $6\cdot 5 \cdot 4 = 120$.
For (c), choose which even digit appears in the ones place ($3$) and then pick the other three numbers in order ($5\cdot 4\cdot 3$).
For (d), the divisibility by $3$ is superfluous, since that happens automatically if the digit sum is $18$. This one just involves some tedious counting. Just start with the big digits and work down.
$(9,9,0,0)$ is one combination. $(9,7,x,x)$ doesn't work, but there is $(9,6,3,0)$ and $(9,3,3,3)$.
For $7$ as the largest digit, we have $(7,7,4,0)$, and that's it.
For $6$ as the largest digit, we have $(6,6,6,0)$, $(6,6,3,3)$, and $(6,4,4,4)$.
There aren't any for $4$ and below, because $4 \cdot 4 = 16 < 18$.
Now, for each of these combinations calculate the number of permutations of the digits, and then add them up. I'll leave that for you. :)