Indian Statistical Institute B.Math & B.Stat : Combinatorics

Using only the digits $2$, $3$ and $9$, how many six digit numbers can be formed which are divisible by $6$? $$$$ First note that a number is divisible by $6$, $iff$ it is divisible by $2$ and by $3.$ ($2$ and $3$ being co-prime)$$$$ The above argument clearly shows that the number $222222$ is divisible by $6$. Use divisibility test of $2$ and $3$. First note that a number will be divisible by $2$, $iff$ the digit at the unit place is $2.$ Now observe that, if the number which is divisible by $2$ has to be divisible by $3$, the sum of the digits must be divisible by $3.$ So the 5 places of the number (except the digit at the unit place, which is $2$) is a combination of the digits $2$, $3$ and $9$ such that the sum including the digit at the unit place is divided by $3$.$$$$ Now observe that among the $5$ places to be filled, exactly $two$ places can be the digit $2$. (Convince yourself why?! thin if not...) And the remaining $three$ places can be filled by either $3$ or $9$. Thus giving \( \binom {5} {2} \times 2\times2\times2 \) possibilities with desired condition. $$$$ therefore, total number of numbers \(= \binom {5} {2} \times 2\times2\times2+1 = 80+1=81 \)