mathmath333
  • mathmath333
Counting question
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
mathmath333
  • mathmath333
|dw:1440718821577:dw|
mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} & \normalsize \text{Find the number of rectangles.}\hspace{.33em}\\~\\ \end{align}}\)

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mathmath333
  • mathmath333
the lines are perpendicular to each other.
ChillOut
  • ChillOut
Begin by choosing two pairs of horizontal lines and two pairs of vertical lines so you can have a rectangle.
anonymous
  • anonymous
I just know theres a ton
ChillOut
  • ChillOut
Let \(\left(\begin{matrix}a \\ 2\end{matrix}\right)\) be the number of vertical lines that can be picked, and \(\left(\begin{matrix}b \\ 2\end{matrix}\right)\) the number of horizontal lines that can be picked. For each pair of horizontal lines, there's a pair of vertical lines that will form a rectangle with them. Therefore you have \(\left(\begin{matrix}n \\ 2\end{matrix}\right)\)*\(\left(\begin{matrix}b \\ 2\end{matrix}\right)\) as the number of rectangles.
ChillOut
  • ChillOut
That would be my guess :)
mathmath333
  • mathmath333
n=4,b=6 ?
ChillOut
  • ChillOut
Yes. And sorry, meant to be "a" in the last line.
mathmath333
  • mathmath333
ok thanks
ChillOut
  • ChillOut
No problem, do you have an answer to the problem?
mathmath333
  • mathmath333
no
mathmath333
  • mathmath333
i m relying on u
ChillOut
  • ChillOut
All right then!

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