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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.

1 year, 4 months ago

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Hint: application of divisibility rules, rule of product Staff · 1 year, 4 months ago

@Sandeep Bhardwaj · 1 year, 4 months ago

@Rishabh Cool · 1 year, 4 months ago

Just one point.. Shouldn't 0 be included since its divisible by 3 also...? · 1 year, 4 months ago

Yeah, just forgot about that one, thanks any other thing that i might have missed. · 1 year, 4 months ago