# Coefficient of $x^7y^6$ in $(xy+x+3y+3)^8$

Find the coefficient of $$x^7y^6$$ in $$(xy+x+3y+3)^8$$.

My solution:

Factor $$(xy+x+3y+3)^8$$ into $$(x+3)^8(y+1)^8$$. To get an $$x^7y^6$$ term, we need to find the coefficient of $$x^7$$ in the first factor and $$y^6$$ in the second factor. Using the binomial theorem, we get the coefficient of $$x^7$$ to be $$17496$$ and $$y^6$$ to be $$28$$. Multiplying the two gets us an answer of $$489888.$$

However, this is wrong. This is the answer key's approach:

$$x^7y^6 = (xy)^6 \cdot x = (xy)^5 \cdot x^2 \cdot y$$. Now, $$(xy)^6\cdot x$$ can be formed by choosing $$6$$ $$xy$$'s, $$1$$ $$x$$, and $$1$$ $$3$$, which can be done in $$\binom{8}{6}\binom{3}{2}\binom{1}{1} = 56$$ ways. $$(xy)^5\cdot x^2\cdot y$$ can be formed by choosing $$5$$ $$xy$$'s, $$2$$ $$x$$'s, and $$1$$ $$3y$$, which can be done in $$\binom{8}{5}\binom{3}{2}\binom{1}{1} = 168$$ ways. Thus the final coefficient is $$3(56+168) = 672$$ ways.

I completely understand their approach, but fail to understand why mine doesn't work. Don't we just calculate the number of ways to get $$x^7$$, and $$y^6$$, then multiply them?

Interestingly enough, I noticed that when you calculate the coefficient of $$x^1$$ (which is $$x^{8-7}$$) and $$y^2$$ (which is $$y^{8-6}$$), you get $$3\cdot \binom{8}{1} \cdot \binom{8}{2} = 672$$, which is the answer. I'm $$99\%$$ sure this isn't a coincidence, but why does this method work and not the other?

I know that I did not get any calculations wrong, because I double-checked everything with WolframAlpha; the error must be in my process.

(Question from PuMaC 2017 Algebra B)

The factor for $$x^7$$ is $$24$$ and for $$y^6$$ is $$28$$, which their multiplication is the correct answer. Probably you're just applying the binomial theorem incorrectly.

• No, I used WolframAlpha just to check and they are correct Commented Aug 17, 2020 at 23:41
• Well, I used Wolfram too, the x one is in fact incorrect, check again for (x+3)^8 Commented Aug 17, 2020 at 23:43
• I just noticed facepalm Commented Aug 17, 2020 at 23:44
• I instead got the $x$ term, not $x^7$ Commented Aug 17, 2020 at 23:45
• to answer more precisely your aproach for the answer is 100% valid Commented Aug 17, 2020 at 23:46