The number of positive integral solutions to the system of equations
$$\begin{align} & a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=47\\
&a_{1}+a_{2}=37,\ \ \{a_{1},a_{2},a_{3},a_{4},a_{5}\} \in \mathbb{N}\end{align}$$is
$a.)\ 2044\quad \quad \quad \quad \quad b.)\ 2246\\
c.)\ 2024\quad \quad \quad \quad \quad \color{green}{d.)\ 2376}$
I know my something like stars and bars for $a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=47$ ,
the non-negative solutions are $\dbinom{47+4}{4}$ and
for $a_{1}+a_{2}=37$
the non-negative solutions are $\dbinom{37+1}{1}$
But the non-negative solutions will include zero and it is not needed here.
Also there are two cases combined I am confused on how to solve this question.
Also this question was given in chapter quadratic equations I don't know how.
I look for a short and simple way.
I have studied maths up to $12$th grade.Thanks.
Best Answer
I think there is no correct option.
The system is equivalent to $$a_1+a_2=37$$ $$a_3+a_4+a_5=10$$
Let $b_i=a_i-1$. Then, $b_i$ are non-negative integers. So, the system is equivalent to $$b_1+b_2=37-2=35$$ $$b_3+b_4+b_5=10-3=7$$ Then, here you can use the method you wrote, so we have $$\binom{36}{1}\times\binom{9}{2}=\color{red}{1296}.$$