TITLE: What is the coefficient of $x^{9}$ in the expansion of $(1+x)^{12}(1+x)^{4}$?
QUESTION [7 upvotes]: 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...

REPLY [3 votes]: If you think about this as choosing objects, the result is actually very intuitive. Picture twelve copies of $(1+x)$ laid out. The only way to get a term of $x^9$ when we expand is to pick exactly $9$ $x$'s and $7$ $1$'s. Alternatively, if we have $12$ copies of $(1+x)$ followed by $4$ more, we can imagine picking the first $k$ $x$'s from the first $12$, and the remaining $9-k$ from the next $4$.