in Combinatory recategorized by
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3 votes
3 votes
How many integer solutions does the equation
$$
x_{1}+x_{2}+x_{3}+x_{4}=15
$$
have, if we require that $x_{1} \geq 2, x_{2} \geq 3, x_{3} \geq 10$ and $x_{4} \geq-3 ?$
in Combinatory recategorized by
177 views

1 Answer

3 votes
3 votes
Let $y_{1}=x_{1}-2, y_{2}=x_{2}-3, y_{3}=x_{3}-10$ and $y_{4}=x_{4}+3$.
We get a new equation where $y_{i} \geq 0$ and
$$
\begin{aligned}
\left(y_{1}+2\right)+\left(y_{2}+3\right)+\left(y_{3}+10\right)+\left(y_{4}-3\right) &=15, \quad \Leftrightarrow \\
y_{1}+y_{2}+y_{3}+y_{4} &=3
\end{aligned}
$$
Dots and bars gives $\left(\begin{array}{c}3+4-1 \\ 4-1\end{array}\right)$ integer solutions.
Answer:

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