# 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...

• It is indeed equal to $(1+x)^{16}$ so your method seems correct to me. – Zubzub Dec 9 '17 at 18:58

You've stumbled across a case of the Vandermonde identity:

$$\sum_{k=0}^r{m\choose k}{n\choose r-k} = {m + n\choose r}$$

So you are indeed correct, the answer is equivalent to the coefficient of $x^9$ in $(1+x)^{16}$, since $${16\choose 9} = \sum_{k=0}^9{4\choose k}{12\choose 9-k}$$ To obtain the second form, note that \begin{aligned} \ [x^9](1+x)^{12}(1+x)^4 & = [x^9]\left(\sum_{k=0}^{12}{12\choose k}x^k\right)\left(\sum_{k=0}^4{4\choose k}x^k\right) \\ & = [x^9]\sum_{k=0}^{12+4}x^k\left(\sum_{i,j\ge0,\ i+j=k}{4\choose i}{12\choose j}\right) \\ & = [x^9]\sum_{k=0}^{16}x^k\left(\sum_{j=1}^{k}{4\choose j}{12\choose k-j}\right) \\ & = \sum_{j=0}^9{4\choose k}{12\choose 9-k} \end{aligned}

Going from the first line to the second line, we've essentially used a finite version of the Cauchy Product, and to go from the inner sum on the second line to the inner sum on the third line, note that if $i+j=k$, then $i=k-j$, so really we only care about $j$.

• oh yes, I have heard of this identity before! Thank you for making it very clear! – numericalorange Dec 9 '17 at 20:24

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 $4$.

• Very nice way to picture it! Thanks! – numericalorange Dec 9 '17 at 20:23

Yes, you can easily put them together to get $(1+x)^{16}$. But, I think that textbook wants you to learn how to split the binomial up. So, in this the reason they have a sum in the answer is because there are many ways to $x^9$. A couple of the ways are $x^0$ from the first term and $x^9$ from the second $x^1$ from the first term, and $x^8$ from the second term, $x^2$ from the first term and $x^7$ from the second term, etc. Summing all of the coefficients up and then compressing it into sigma notation creates their answer. For me, I would obviously do it your way, but I think the book wants you to take something away from it :D

The two answers are in fact equal. Check out the Vandermonde identity.

• Oh, I understand why they decided to do that then! Thanks! – numericalorange Dec 9 '17 at 20:23

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!!!

• Oh, I understand! Thank you!! – numericalorange Dec 9 '17 at 20:23