Coefficient of $y^3.x^6$ in $(1+x+y)^5 . (1+x)^7$

Question

I actually just expanded it and got the answer: $840$.

But I am pretty sure that there`s some binomial (tri-nomial?) expansion involved here that can lead to the answer much more efficiently.

I did try to expand the binomial using the theorem and picked the $x^6$ term, and then expand the trinomial, but it became a mess because getting $y^3.x^6$ seems to involve more than just one term from the $(1+x)^7$.

All in all, can anyone shed any light on this?

Answer 1

You have $$(1+x+y)^5(1+x)^7=\sum_{k=0}^5\binom5k(1+x)^{12-k}y^k.$$ The coefficient of $x^6y^3$ herein is the coefficient of $x^6$ in $\binom53(1+x)^9$, that is $\binom53\binom96$.