- #1
tylerc1991
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Homework Statement
What is the coefficient on [itex]x^{12} y^{24}[/itex] in [itex](x^3 + 2xy^2 + y + 3)^{18}[/itex]?
Homework Equations
Multinomial Theorem:
[itex]\displaystyle \left( \sum_{k = 1}^m x_k \right)^n = \sum_{k_1 + \dotsb + k_m = n} \binom{n}{k_1, \dotsc, k_m} \prod_{i = 1}^m x_i^{k_i}[/itex]
The Attempt at a Solution
Using the multinomial theorem, the expansion of [itex](x^3 + 2xy^2 + y + 3)^{18}[/itex] has terms of the form
[itex]\binom{18}{k_1, k_2, k_3, k_4} (x^3)^{k_1} (2xy^2)^{k_2} (y)^{k_3} (3)^{k_4} = \binom{18}{k_1, k_2, k_3, k_4} x^{3k_1} 2^{k_2} x^{k_2} y^{2k_2} y^{k_3} 3^{k_4} = \binom{18}{k_1, k_2, k_3, k_4} x^{3k_1 + k_2} 2^{k_2} y^{2k_2 + k_3} 3^{k_4}.[/itex]
The [itex]x^{12} y^{24}[/itex] occurs when [itex]3k_1 + k_2 = 12[/itex] and [itex]2k_2 + k_3 = 24[/itex]. We also know that [itex]k_1 + k_2 + k_3 + k_4 = 18[/itex]. This is where I get stuck, since I have 3 equations and 4 unknowns. Is my work correct up to this point? Is there another equation that I can use to find each [itex]k_i[/itex]? Thank you!