How many numbers with 6 digits can be formed with the digits $1,2,3,4,5$ such that the digit $2$ appears every time at least three times.
Total numbers: $5^6$
Numbers in which 2 doesn't appear: $4^6$
Numbers in which 2 appear once : $6\cdot4^5$
Numbers in which 2 appear twice : $13\cdot4^4$
So my result is: $5^6-4^6- 6 \cdot4^5-13\cdot4^4=2057 $ but the right answer is $1545$ How solve it ?