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.

Thanks in advance!

(Question from PuMaC 2017 Algebra B)


1 Answer 1


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.

As an add-on, your method is completely correct.

  • $\begingroup$ No, I used WolframAlpha just to check and they are correct $\endgroup$ Commented Aug 17, 2020 at 23:41
  • $\begingroup$ Well, I used Wolfram too, the x one is in fact incorrect, check again for (x+3)^8 $\endgroup$
    – aangulog
    Commented Aug 17, 2020 at 23:43
  • $\begingroup$ I just noticed facepalm $\endgroup$ Commented Aug 17, 2020 at 23:44
  • 1
    $\begingroup$ I instead got the $x$ term, not $x^7$ $\endgroup$ Commented Aug 17, 2020 at 23:45
  • $\begingroup$ to answer more precisely your aproach for the answer is 100% valid $\endgroup$
    – aangulog
    Commented Aug 17, 2020 at 23:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .