Newton Binomial to find $(x_1+x_2+x_3)(x_4+x_5+x_6+x_7)=77$

Question

What is the number of solutions to $(x_1+x_2+x_3)(x_4+x_5+x_6+x_7)=77$, where $x_1,x_2,\dots,x_6,x_7$ are non negative integers? Obviously the $2$ multipliers are $1 \times 77$ and $7 \times 11$, so we have $x_1+x_2+x_3 = 1,7,11,77$ and $x_4+x_5+x_6+x_7 = 1,7,11,77$.

Answer 1

Recall the method of "stars and bars," which tells us that the number of nonnegative integer solutions to the equation $$ x_1 + x_2 + \cdots + x_k = n $$ is $$ \binom{n+k-1}{n}. $$

There are four cases, as you pointed out. The tuple $(x_1 + x_2 + x_3, x_4 + x_5 + x_6 + x_7)$ can only equal $(1, 77)$, $(7, 11)$, $(11, 7)$, or $(77, 1)$. In the first case, where $x_1 + x_2 + x_3 = 1$ and $x_4+x_5+x_6+x_7=77$, the number of solutions is $$ \binom{3}{1}\binom{80}{77}. $$

Applying this reasoning for the other three cases, we get that the total number of solutions is $$ \binom{3}{1}\binom{80}{77} + \binom{9}{7}\binom{14}{11} + \binom{13}{11}\binom{10}{7} + \binom{79}{77}\binom{4}{1}. $$