How many exchanges are needed to serve a city of 80,000 people?

AI Thread Summary
A seven-digit phone number in the U.S. includes a three-digit exchange and a four-digit number, allowing for a maximum of 10,000 unique numbers per exchange. To serve a city of 80,000 people, at least eight exchanges would be necessary, as each exchange can accommodate 10,000 numbers. The discussion highlights the use of combination, permutation, and arrangement equations to understand these calculations. The first part of the question confirms that an area code can support up to 8 million numbers, provided the first digit of the exchange is not 0 or 1. Understanding these principles is crucial for determining the number of exchanges required for a specific population.
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A seven-digit phone number in the United States consists of a three-digit exchange followed by a four-digit number. How many exchanges are needed to serve a city of 80,000 people?


Combination, Permutation, and arrangement with repetition equations are used in this section.

Part 1 of this question I got right: 8,000,000 people can have phone numbers in one area code if the first digit of the 7 numbers can't be 0 or 1. Part 2 listed above I'm absolutely stumped on though.
 
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10,000 numbers per exchange?
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
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