The total number of six digit numbers, formed using the digits 4, 5, 9 only and divisible by 6, is _______

We have to form 6 digit number using 4, 5, 9 and should be divisible by 6 which means that the addition of number should be divisible by 3 and unit place should be even. 
C₁ : The number is 4,4,4,4,4,4 which can be arranged in one way. 
C₂ : Number is joined using the digits 4,4,4,4,5,9 and last place is fixed with 4 that can be arranged in $\frac{5!}{3!}=20 \mathrm{ways}$
C₃ : When number is formed using the digits 4,4,5,5,5 and last place fixed with 4 that can be arranged in $\frac{5!}{2!3!}=10 \mathrm{ways}$
C₄ : Number is formed using the digits 4,4,4,9,9,9 and last place fixed with 4 that can be arranged in $\frac{5!}{2!3!}=10 \mathrm{ways}$
C₅ : Number is formed using the digits 4,4,5,5,9,9 and last place fixed with 4 that can be arranged in $\frac{5!}{2!2!}=30 \mathrm{ways}$
C₆ : Number is formed using the digit 4,5,9,9,9,9 and last digit fixed with 4 that can be arranged in $\frac{5!}{4!}=5 \mathrm{ways}$
C₇ : Number is formed using the digit 4,5,5,5,5,9 and last digit fixed with 4 that can be arranged in $\frac{5!}{4!}=5 \mathrm{ways}$
$\therefore$Required = 1 + 20 + 10 + 10 + 30 + 5 + 5 = 81