<s>[INST] $4$-digit positive integers that does not contain the digits $3$ and $4$ plus other properties I'm just wondering, what is the total number of $4$ digit positive integers that have the following properties:
All digits are unique.
Does not contain the digits $3$ and/or $4.$
The number is divisible by $3.$
Thoughts:  All digits are unique is $9\times9\times8\times7,$ as I have learned, and I don't know how to do the second and third ones, to eliminate the ones that doesn't have this property.
Thanks!
 [/INST] Let $A_0=\{6,9\}$, $A_1=\{1,7\}$, and $A_2=\{2,5,8\}$.
1) If 0 is not a digit in the number, then the digits mod 3 are equal to
   $\{0,0,1,2\}$, $\{0,2,2,2\}$, or $\{1,1,2,2\}$  since an integer is divisible by 3 iff the sum of its digits is divisible by 3.
Therefore there are $\binom{2}{2}\binom{2}{1}\binom{3}{1}+\binom{2}{1}\binom{3}{3}+\binom{2}{2}\binom{3}{2}$ ways to choose the digits, and there are 4! ways to arrange them.
2) If 0 is a digit in the number, then the other three digits mod 3 are equal to $\{0,1,2\}$ or $\{2,2,2\}$.
Therefore there are $\binom{2}{1}\binom{2}{1}\binom{3}{1}+\binom{3}{3}$ ways to choose the digits, and $3(3!)$ ways to arrange them.
Thus there are $11\cdot24+13\cdot18=498$  such numbers.
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