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I was given the next question: How many $5$ digit numbers are there that have $3$ as one of their digits and are also divided by $3$?

I'd like to hear your thoughts on my answer or suggest a better one:

My answer would be finding the number of all of the possible $5$ digit numbers and then reduce it by $(B+C+D)$ whereas:

$B$= all of the possible options for a $5$ digit number which is divided by $3$ but does not contain the digit $3$ in it other than the last digit.

$C$=all of the possible options for a $5$ digit number which is not divided by $3$ but does contain the digit $3$ in it.

$D$=all of the possible options for a $5$ digit number which is not divided by $3$ and does not contain the digit $3$ in it.

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    Do you know PIE (inclusion-exclusion principle)? If so another solution would be calculating the number of 4 digits numbers divisible by 3 and how often can you insert a 3 at any place, then subtract the numbers you counted double (these with two 3s) and so on... – 2017-01-06
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    @ctst I haven't heard of it..is it taught at an early stage? – 2017-01-06
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    Yes. The two-set principle of inclusion exclusion for set theory is stated as $|A\cup B|=|A|+|B|-|A\cap B|$. Using this, one can form the general principle of inclusion-exclusion for arbitrary number of sets inductively. – 2017-01-06
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    @Lola Mostly somewhere in the middle of a combinatorics course (depending on your professor and if it is a pure combinatorics course). You can find it here: https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle Just look at the statement and consider your sets as $A_i:=\{$has as $i$th digit a 3$\}$ – 2017-01-06

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