Find the number of six-digit numbers, all of whose digits belong to \1,2,3,4\, in which each of the digits 1,2,3,4 appears at least once and which…

A number is divisible by $4$ exactly when its last two digits form a multiple of $4$. With digits from $\{1,2,3,4\}$ the admissible endings are $12,\ 24,\ 32,\ 44$. For each ending, the first four digits must contain every digit missing from the ending. Ending $12$ (missing $3,4$): strings of length 4 over 4 digits containing both $3$ and $4$ number $4^4-2\cdot3^4+2^4=256-162+16=110$. The endings $24$ (missing $1,3$) and $32$ (missing $1,4$) give $110$ each by the same count. Ending $44$ (missing $1,2,3$): $4^4-3\cdot3^4+3\cdot2^4-1^4=256-243+48-1=60$. The total is $110+110+110+60=390$. A brute-force enumeration of all $4^6$ strings gives $390$.