What is the coefficient of $x^{11}$ in $(3x-9)^{19}$?

Question

I am currently studying for finals, and I do not know how to do this problem from my study guide. I have tried to watch a few YouTube videos and I know that I will end up with $3x^{11} \times (-9)^8$, but from here I do not know where to go. The YouTube videos I have watched have been unclear, and I do not have a chance to ask my professor until just a little before the exam. If anyone has any tips on where to go from here I would greatly appreciate it!

Answer 1

$$(3x-9)^{19} = (3x-9)(3x-9) \cdots (3x-9) $$

Notice that on the right hand side there exist $19$ products. A term can be formed by choosing either $3x$ or $-9$ from each of these $19$ products.

If your goal is to find the coefficient of $x^{11}$, you have to choose $3x$ for a $11$ times from the available $19$ products, and then choose $-9$ for $8$ times from the remaining $8$ products.

If you choose $3x$ for $11$ times, you end up with $\color{blue}{(3x)^{11}}$, and you can do this in exactly $\color{green}{\binom{19}{11}}$ ways. $\color{grey}{\text{(because you have $19$ products available and you're considering only $11$ of them for $3x$.)}}$

For each of the above $\color{green}{\binom{19}{11}}$ ways, you can choose $-9$ from the remaining $8$ products, this gives $\color{red}{(-9)^8}$
So the overall term would be $\color{green}{\binom{19}{11}}\color{blue}{(3x)^{11}}\color{red}{(-9)^8}$

Answer 2

Note that by expansion of a binomial to the $n^{th}$ power, you get

$$(a+b)^n = \sum_{r = 0}^{n} {n \choose r}a^{n-r}b^r = a^n+a^{n-1}b+a^{n-2}b^2+…+a^{n-r}b^r+…+ab^{n-1}+b^n$$

where ${n \choose r}$ corresponds to the binomial coefficients, or Pascal Numbers. (Never forget those for coefficients.)

For $(3x-9)^{19}$, the first term has $x^{19}$, the second term has $x^{18}$, etc. So, $x^{11}$ corresponds to the ninth term. Hence, you get

$${19 \choose 19-8} (3x)^{11}(-9)^8$$

$${19 \choose 11} (3x)^{11}9^8$$

Simplifying from here, you can solve for the ninth term and find the coefficient of $x^{11}$.

Answer 3

HINT: The Binomial expansion : $$(a+b)^{n}= \sum_{k=0}^{n} {n \choose k}{a}^{k}{b}^{n-k}.$$

Substitute $a=3x$ and $b=-9$ . To find out the coefficient of $x^{11} $,put $k=11$ in ${n \choose k}{a}^{k}{b}^{n-k}$

Answer 4

You are part way there.

If you were to write out $(3x-9)^{19}$ in full, it would be a product of nineteen $ (3x-9)$. To get an $x^{11}$ you would need to pick 11 of these $(3x-9)$.

There are $19 \choose 11$ ways of doing that, so the term in $x^{11}$ is $$ {19 \choose 11} (3x)^{11} (-9)^8 $$

Thus the coefficient of $x^{11}$ is

$$ {19 \choose 11} (3)^{11} (-9)^8 = {19 \choose 11}3^{27} $$