Probability Part 1

Given, digits $=\{0,1,2,3,4,5,6\}$ Number of ways in which 6 digit number can be formed using these 7 digits $=6 \cdot 6!$ Number of ways ⇒ $\begin{array}{cccccc} \overline{\downarrow} & \overline{\downarrow} & \overline{\downarrow} & \overline{\downarrow} & \overline{\downarrow} & \overline{\downarrow} \\ 6 & 6 & 5 & 4 & 3 & 2 \end{array}$ ( 0 can't be filled here) If the required number is divisible by 3 , then the sum of the digits must be divisible by 3 . Sum of all 7 digits $=0+1+2+3+4+5+6$ $=\frac{6 \cdot 7}{2}=21$ Now, this further implies we have to remove anyone digit from the given digits. We can remove only the multiple of 3 as the sum of all 7 digits is already 21 . So, removing any other digit will lead to a number of 6 digit which will not be a multiple of 3 . Possible digits $\Rightarrow \{1,2,3,4,5,6\}$ $\{0,1,2,4,5,6\}$ $\{0,1,2,3,4,5\}$ Case I $\{1,2,3,4,5,6\}$ Number of 6 digit numbers $=6!$ Case II $\{0,1,2,4,5,6\}$ Case III $\{0,1,2,3,4,5\}$ Number of 6 digit numbers $=5 \cdot 5!$ $\therefore$ Probability $=\left(\frac{6!+5 \cdot 5!+5 \cdot 5!}{6 \cdot 6!}\right)$ $=\left(\frac{1}{6}+\frac{5}{36}+\frac{5}{36}\right)=\frac{16}{36}=\frac{4}{9}$