Binomial expansion and combinatorics

Question

Coefficient of $x^{11}$ in the expansion of $(1+x^2)^4 (1+x^3)^7 (1+x^4)^{12}$ is

If I were to write out all the product terms straight out, I would have 23 terms to choose from. How do I actually apply such combinatorics approach to solve it? I realize that I will have to sum up different cases of choosing out terms to make up $x^{11}$ but can someone help me figure out how?

Answer 1

Remember the binomial theorem tells us that the coefficient of $a^k$ in $(1+a)^n$ is $_nC_k$.

Identify the different compositions of terms where the product gives $x^{11}$.

From $(1+x^2)^4(1+x^3)^7(1+x^4)^{12}$, we have $x^0x^3x^8,x^2x^9x^0,x^4x^3x^4,x^8x^3x^0$. If you find the coefficients for each of these, and add them up, you'll get the answer (which should be $1113$).

Answer 2

Put $P(u)=(1+u)⁴(1+u^2)^{12}$ and $Q(u)=(1+u)^7$. your expression is $P(x^2)Q(x^3)$. You obtain a term in $x^{11}$ by the product of a term in $P(x^2)$, say $x^{2k}$ (up to a constant), and a term in $Q(x^3)$, say $x^{3l}$, we must have $11=2k+3l$. Compute the solutions to this equation, , $k=1$ and $l=3$ and $k=4$, $l=1$. So you have to find the coefficient of $u^3$ in $Q$, and the coefficient of $u$ in $P(u)$ (ie to compute $P^{\prime}(0)$), and to multiply them. Do the same for the second case. It is easy to finish.