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

$$[y^3] (1+x+y)^5 (1+x)^7 = [y^3](z+y)^5 z^7= {5 \choose 3} z^9.$$ Then the required co-efficient of $x^6 y^3~ in~ (1+x+y)^5 (1+x)^7$ is $${5 \choose 3}[x^6](1+x)^9= {5 \choose 3} {9 \choose 6}.$$

Answer 2

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 $\binom53\binom96$.

Answer 3

Using the binomial theorem on $(1 + x + y)^5$, we have:

$$\left( (1+x)^5 + {5 \choose 1}(1+x)^4y + {5 \choose 2}(1+x)^3y^2 + {5 \choose 3}(1+x)^2y^3 + {5 \choose 4}(1+x)y^4 + {5 \choose 5}y^5 \right) \cdot(1+x)^7$$

which has the same coefficient of $y^3 \cdot x^6$ as:

$${5 \choose 3}(1+x)^2 y^3 \cdot (1+x)^7$$

Then split into cases. When the coefficient of left bracket is $x^0$ and the right bracket is $x^7$, the coefficient of $y^3 \cdot x^6$ is ${5 \choose 3} \times {7 \choose 6}$. Doing this for the other cases gives a total of:

$${5 \choose 3} \left(1 \times {7 \choose 6} + 2 \times {7 \choose 5} + 1 \times {7 \choose 4} \right)$$ $$=840$$