Multinomial Theorem Problem

[tylerc1991]

Homework Statement

What is the coefficient on $x^{12} y^{24}$ in $(x^3 + 2xy^2 + y + 3)^{18}$?

Homework Equations

Multinomial Theorem:
$\displaystyle \left( \sum_{k = 1}^m x_k \right)^n = \sum_{k_1 + \dotsb + k_m = n} \binom{n}{k_1, \dotsc, k_m} \prod_{i = 1}^m x_i^{k_i}$

The Attempt at a Solution

Using the multinomial theorem, the expansion of $(x^3 + 2xy^2 + y + 3)^{18}$ has terms of the form

$\binom{18}{k_1, k_2, k_3, k_4} (x^3)^{k_1} (2xy^2)^{k_2} (y)^{k_3} (3)^{k_4} = \binom{18}{k_1, k_2, k_3, k_4} x^{3k_1} 2^{k_2} x^{k_2} y^{2k_2} y^{k_3} 3^{k_4} = \binom{18}{k_1, k_2, k_3, k_4} x^{3k_1 + k_2} 2^{k_2} y^{2k_2 + k_3} 3^{k_4}.$

The $x^{12} y^{24}$ occurs when $3k_1 + k_2 = 12$ and $2k_2 + k_3 = 24$. We also know that $k_1 + k_2 + k_3 + k_4 = 18$. 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 $k_i$? Thank you!

 

[mighty2000]

Yes, your work is correct up to this point. You are correct in saying that you have 3 equations and 4 unknowns, so you cannot solve for each k_i individually. However, you can use the fact that k_1, k_2, k_3, and k_4 are all non-negative integers to narrow down the possible values of k_i.

First, we know that k_2 must be even in order for 2^{k_2} to be a perfect square. This means that k_2 can only take on the values 0, 2, 4, 6, ..., 18.

Next, we can use the equation 3k_1 + k_2 = 12 to solve for k_1 in terms of k_2. We get k_1 = (12 - k_2)/3. Since k_1 must be a non-negative integer, this means that k_2 must be a multiple of 3.

Finally, we can use the equation 2k_2 + k_3 = 24 to solve for k_3 in terms of k_2. We get k_3 = 24 - 2k_2. Again, since k_3 must be a non-negative integer, this means that k_2 must be less than or equal to 12.

Using this information, we can now list out all the possible values of k_1, k_2, and k_3 that satisfy the given equations:

k_1 = 0, k_2 = 0, k_3 = 24
k_1 = 0, k_2 = 3, k_3 = 18
k_1 = 0, k_2 = 6, k_3 = 12
k_1 = 0, k_2 = 9, k_3 = 6
k_1 = 0, k_2 = 12, k_3 = 0
k_1 = 1, k_2 = 0, k_3 = 21
k_1 = 1, k_2 = 3, k_3 = 15
k_1 = 1, k_2 = 6, k_3 = 9
k_1 = 1, k_2 =