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Find the no of ways of placing 6 identical balls into 3 distinct boxes in such a way that first box contains 0,1 or 2 objects,2bd box contains 1,2,3 objects and 3rd one contains 3 or 5 objects.

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I tried using combinatorial problem

That I find 6 !

And after 6C2 ,6C3

I can't get answer. Please help.

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I don't know if I am correct but can the answer be the coefficient of $x^6$ in the expansion of

$$(1+x+x^2)(x+x^2+x^3)(x^3+x^5)$$

Won't be wrong if I say that right?

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  • $\begingroup$ Please explain how to fimd $\endgroup$ – Akash Sarkar Feb 21 at 5:33
  • $\begingroup$ Yup, that's right. In this scenario, we can use generating functions to do it. Might be a little pain to multiply out the polynomials, but oh well, that's how these problems tend to go. $\endgroup$ – Eevee Trainer Feb 21 at 5:34
  • $\begingroup$ @Akash_Sarkar Just expand $$x^4(1+x+x^2)^2(1+x^2)=x^4(1+x^2+x^4+2x+2x^3+2x^2)(1+x^2)$$. And I don't think there would be a problem to find coefficient of $x^2$ in $(1+x^2+x^4+2x+2x^3+2x^2)(1+x^2)$ $\endgroup$ – Darkrai Feb 21 at 5:35
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Let $(a,b,c)$ represent the number of objects in 3 boxes in that order. That is, $(a,b,c)$ means that the 1st box has $a$ balls, the 2nd box has $b$ balls, and so on.

Then all possible cases are $$ (0,1,5), (0,3,3), (1,2,3), (2,1,3) $$ Therefore, the total number of ways of placing 6 balls is $$ \binom{6}{0,1,5} + \binom{6}{0,3,3} + \binom{6}{1,2,3} + \binom{6}{2,1,3}. $$

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