Find the total number of ways of formation of numbers, which are divisible by 3, utilizing 0,1,2,3,4 and 5 without repetition.

Well the question that i was solving asked to find out the number of ways of forming five digit numbers and since I had put a lot of effort in solving it just to find out that i had misread the question, I posted the question here, so as to verify whether my approach is correct or not. Btw I am a noob at combinatorics.

So what I did was as follows:-

1 digit numbers:- 1(i.e. the number 3)

2 digit numbers:-

Lets consider that the number is formed from two digits \({ x }_{ 1 }\) & \({ x }_{ 2 }\) and as the number should be a multiple of 3 so,

\({ x }_{ 1 }+{ x }_{ 2 }=3n\), where \(n=1,2,3; 0{ <x }_{ 1 }\le 5; 0{ \le x }_{ 1 }\le 5\)

Now, using multinomial expansion, we get \((x+{ x }^{ 2 }+{ x }^{ 3 }+{ x }^{ 4 }+{ x }^{ 5 })(1+x+{ x }^{ 2 }+{ x }^{ 3 }+{ x }^{ 4 }+{ x }^{ 5 })\) where we have to find the coefficient of \({ x }^{ 3n }, n=1,2,3\)

And proceeding forward like this till the six digit numbers. Please do point out any other conditions that i might not have considered here and which would be required in further cases.

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## Comments

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TopNewestHint:application of divisibility rules, rule of productLog in to reply

@Sandeep Bhardwaj

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@Rishabh Cool

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Just one point.. Shouldn't 0 be included since its divisible by 3 also...?

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Yeah, just forgot about that one, thanks any other thing that i might have missed.

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