The total number of six digit numbers, formed using the digits 4,...

Given,

We have to form $6$ digit number using the digits $4, 5 \text{ and } 9$ and for number should be divisible by $6$ which means that the addition of number should be divisible by $3$ and unit place should be even, so we will make following cases:

Case $\left(1\right)$ The number is $444444$ which can be arranged in $1$ way,

Case $\left(2\right)$ when the number is formed using the digits $4, 4, 4, 5, 9$ and last place is fixed with $4$, that can be arranged in $\dfrac{5 !}{3 !} = 20\text{ways}$

Case $\left(3\right)$ when number is formed using the digits $4, 4, 5, 5, 5$ and last place is fixed with $4$, that can be arranged in $\dfrac{5 !}{3 ! 2 !} = 10\text{ways}$

Case $\left(4\right)$ when number is formed using the digits $4, 4, 9, 9, 9$ and last place is fixed by $4$, that can be arranged in $\dfrac{5 !}{3 ! 2 !} = 10\text{ways}$

Case $\left(5\right)$ when number is formed using the digits $4, 5, 5, 9, 9$ and last place is fixed with $4$, that can be arranged in $\dfrac{5 !}{2 ! 2 !} = 30\text{ways}$

Case $\left(6\right)$ when number is formed using the digits $5, 9, 9, 9, 9$ and last place is fixed with $4$, that can be arranged in $\dfrac{5 !}{4 !} = 5\text{ways}$

Case $\left(7\right)$ when number is formed using the digits $5, 5, 5, 5, 9$ and last place is fixed with $4$, that can be arranged in $\dfrac{5 !}{4 !} = 5\text{ways}$

Now adding all the cases we get, $1 + 20 + 10 + 10 + 30 + 5 + 5 = 81\text{ways}$.