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How many numbers of 6 digits, that can be formed with digits 2,3,9. And also divided by 3? $$$$ I was trying to to add 2^6 (when there is no 2)+ C(6,2)2^2 (when 2 can be in two places)+C(6,3)2^3 (when 2 can be in three places)...

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  • $\begingroup$ Closely related to this and this. $\endgroup$ – Lucian Dec 28 '13 at 17:45
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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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  • $\begingroup$ Why 2 can't appear 2 times? $\endgroup$ – user7777777 Dec 28 '13 at 8:01
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    $\begingroup$ @user7777777 Because then the digit sum isn't divisible by $3$. $\endgroup$ – user61527 Dec 28 '13 at 8:01
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    $\begingroup$ Shouldn't it be 2^3 instead of 3^2? $\endgroup$ – sebii Dec 28 '13 at 8:06
  • $\begingroup$ @sebii Yes, thanks. $\endgroup$ – user61527 Dec 28 '13 at 8:09
  • $\begingroup$ instead of being divided by 3, now divided by 6? how can I solve this? $\endgroup$ – user7777777 Dec 28 '13 at 8:55
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The number of twos have to be zero, three or six.

So your choices are

1) No twos and the six digits use 3's and 9's

2) Three twos, and the remaining six digits use 3's and 9's.

3) All 6 are twos

I think you know how to work out the number of each case

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