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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.

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  • 1
    $\begingroup$ Consider that $a_3 + a_4 + a_5 = 10$. $\endgroup$ – Alex Zorn Jul 22 '15 at 21:36
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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 $$\binom{36}{1}\times\binom{9}{2}=\color{red}{1296}.$$

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Hint: You can look up "stars and bars" to find the formula for positive integer solutions (or just use the formula for non-negative solutions, by adding 1 automatically to each number and then subtracting the number of numbers from the sum, and using your formula for non-negative solutions for the lowered sum). Then, you have a solution if and only if $a_1 + a_2 = 37$ and $a_3 + a_4 + a_5 = 10$, and these events are independent.

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  • $\begingroup$ u mean $38+\dbinom{12}{2}$ ?, didn't understand ur hint. $\endgroup$ – R K Jul 22 '15 at 21:42
  • $\begingroup$ @RK I'm not sure if I understand your question, but basically what I'm saying is that the number of positive integer solutions, for example, of $a_1 + a_2 + a_3 + a_4 + a_5 = 47$ is equal to the number of non-negative solutions to $a_1 + a_2 + a_3 + a_4 + a_5 = 47 - 5 = 42$. $\endgroup$ – user2566092 Jul 22 '15 at 21:47
  • $\begingroup$ @RK Also, I should have been more clear, by "independent" I mean you can multiply the number of solutions for $a_1 + a_2$ with the number of solutions for $a_3 + a_4 + a_5$. $\endgroup$ – user2566092 Jul 22 '15 at 21:48
  • $\begingroup$ is this correct $\dbinom{46}{4}\times \dbinom{36}{1}$ ??? $\endgroup$ – R K Jul 22 '15 at 21:57
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$$a_1+a_2+a_3+a_4+a_5=47\\a_1+a_2=37\\ \left\{\begin{matrix} a_3+a_4+a_5=47-37\\ a_1+a_2=37 \end{matrix}\right. $$ multiply those answers $$ \binom{37+2-1}{2-1}*\binom{10+3-1}{3-1}\\=\binom{38}{1}*\binom{12}{2}$$

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