Method Of Fictious Partition Problem

I read in a book that the number of solutions of x+y+z=11 where x,y,z belong to [1,6] and are integers, can be given by coefficient of \(x^{11}\) in expansion of \((x+x^2+x^3+.....+x^6)^3\). I can't understand the method. Can someone explain the logic behind it? And is there a general statement for this?

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Note by Shubham Srivastava
7 years, 10 months ago

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Comments

Consider using the distributive property to write the following product as the sum of a bunch of terms:

(t1+t2+t3+t4+t5+t6)(t1+t2+t3+t4+t5+t6)(t1+t2+t3+t4+t5+t6)(t^1+t^2+t^3+t^4+t^5+t^6)(t^1+t^2+t^3+t^4+t^5+t^6)(t^1+t^2+t^3+t^4+t^5+t^6).

If you pick the txt^x term from the first sum, the tyt^y term from the first sum, and the tzt^z term from the first sum, then the product will be tx+y+zt^{x+y+z}.

So, each solution (x,y,z)(x,y,z) to x+y+z=11x+y+z = 11 with x,y,z∈1,6‾x,y,z \in \overline{1,6} yields a t11t^{11} term.

Thus, the t11t^{11} coefficient of the expansion is the number of such solutions.

This method goes under the category of Generating Functions. There are several sources online that formalize this and give more examples.

Jimmy Kariznov - 7 years, 10 months ago

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Oh! It was so simple. Thanks for explaining.

Shubham Srivastava - 7 years, 10 months ago

I am also not able to understand this theorem.I am also stuck with it.I think here multinomial theorem is used.

Kishan k - 7 years, 10 months ago
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