<s>[INST] What is the coefficient of $x^{9}$ in the expansion of $(1+x)^{12}(1+x)^{4}$? Answer in the textbook: $\sum_{k=0}^{4} \binom{12}{9-k}\binom{4}{k}$
Can I just multiply the two terms together to get $(1+x)^{16}$ then apply the binomial coefficient theorem?
I don't know how they got the sum at the end, I was only taught how to find the coefficient of a term in one binomial expansion using the binomial coefficient theorem...
 [/INST] The factorization $(1+x)^{12}(1+x)^4$ is not accidental here. Let's expand by using Binomial Theorem:
$$\sum_{k=0}^{12}\binom{12}{k}x^k\cdot\sum_{m=0}^{4}\binom{4}{m}x^m.$$
What about $x^9$ in this expansion? We get this power when $k=9,m=0$, $k=8,m=1$, $k=7,m=2$, $k=6,m=3$ and finally $k=5,m=4$. Then our coefficient is $$\binom{12}{9}\cdot\binom{4}{0}+\binom{12}{8}\cdot\binom{4}{1}+\binom{12}{7}\cdot\binom{4}{2}+\binom{12}{6}\cdot\binom{4}{3}+\binom{12}{5}\cdot\binom{4}{4}.$$ It's that!!!
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