Multinomial Theorem: Find Coefficient of x^{12}y^{24}

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SUMMARY

The discussion focuses on finding the coefficient of the term \(x^{12}y^{24}\) in the expression \((x^3 + 2xy^2 + y + 3)^{18}\) using the multinomial theorem. The user sets \(x_1 = x^3\), \(x_2 = 2xy^2\), \(x_3 = y\), and \(x_4 = 3\), leading to the equations \(3k_1 + k_2 = 12\), \(2k_2 + k_3 = 24\), and \(k_1 + k_2 + k_3 + k_4 = 18\). The user expresses confusion regarding negative values for \(k_4\) and is reminded that all \(k_i\) must be nonnegative, indicating a need for careful consideration of the constraints on \(k_1\), \(k_2\), \(k_3\), and \(k_4\).

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Homework Statement


Find the coefficient of the x^{12}y^{24} for (x^3+2xy^2+y+3)^{18}.


Homework Equations


Multinomial theorem, as stated on http://en.wikipedia.org/wiki/Multinomial_theorem


The Attempt at a Solution


Using the multinomial theorem in the form of the wikipedia post, I would set x_1=x^3, x_2=2xy^2, x_3=y, x_4=3. Now I need to find the k's that "match" the coefficients of the term from the problem statement. This will give me a relation between the k's of the following form
3k_1+k_2 = 12, 2k_2+k_3 = 24, and k_1+k_2+k_3+k_4=18. Now I let k_1 vary and record the discrete values k_2, k_3, k_4, but what I'm confused on is why do I sometimes get negative values for k_4? Do I need to do something different with the relationship of the k's. Thanks in advance.
 
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You wrote down three restrictions, but there is a fourth as well, namely that all the k_i's must be nonnegative. Looks like k_1 can be 0 or 1, and that's all.
 
cool, so everything else looks good?
 

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