How many numbers of 6 digits, that can be formed with digits 2,3,9. And also divided by 3?

combinatorics

Solution 1:

For the first question, you have three choices for each of the six digits.

As a hint for the second question: a number is divisible by $3$ if and only if its digit sum is divisible by $3$. If your number is formed out of $2$'s, $3$'s and $9$'s, this imposes conditions on how many $2$'s can appear: In particular, $2$ must appear $0$, $3$ or $6$ times.

As a particular case, for when it appears $3$ times: There are $6 \choose 3$ options for where to place the twos, and each of the remaining three positions have two choices; this leads to

$$2^3 \cdot {6 \choose 3}$$

possibilities.

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