Mathematics - Permutation & Combination Question with Solution | TestHub

Given,

We have to form $6$ digit number using the digits $4, 5 \& 9$ 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 $\frac{5!}{3!}=20 \text{ways}$

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 $\frac{5!}{3!2!}=10 \text{ways}$

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 $\frac{5!}{3!2!}=10 \text{ways}$

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 $\frac{5!}{2!2!}=30 \text{ways}$

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 $\frac{5!}{4!}=5 \text{ways}$

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 $\frac{5!}{4!}=5 \text{ways}$

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