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 April 1st, 2014, 06:20 PM #1 Newbie   Joined: Apr 2014 From: Auckland Posts: 3 Thanks: 0 Telephone Network Hi, Please help me with the following, im unsure how to write up the equation. Problem: Imagine a network of Telephones. There are 10 Telephones in the Network. Each Telephone is connected with a cable to each other Telephone. How many cables are there? And How many cables would you require for a network of 20 Telephones? Thanks.
 April 1st, 2014, 06:48 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra Start with one of the telephones. Count the cables you need to link it to the rest. Do the same for the second telephone bearing in mind that it is already linked to the first. Repeat for all telephones.
April 1st, 2014, 07:48 PM   #3
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Quote:
 Originally Posted by AU19852 Hi, Please help me with the following, im unsure how to write up the equation. Problem: Imagine a network of Telephones. There are 10 Telephones in the Network. Each Telephone is connected with a cable to each other Telephone. How many cables are there? And How many cables would you require for a network of 20 Telephones? Thanks.
You mean that there's cable connecting every single telephone? That is really inefficient...

Are you familiar with binomial coefficients? Since a cable connects two phones, and there are 10 of them, then there are 10C2 = 45 cables in total. With 20 phones, there are 20C2 cables in total.

April 3rd, 2014, 06:08 AM   #4
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 Originally Posted by AU19852 Problem: Imagine a network of Telephones. There are 10 Telephones in the Network. Each Telephone is connected with a cable to each other Telephone. How many cables are there? And How many cables would you require for a network of 20 Telephones?
The easiest way to solve this problem is finding the number of diagonals in polygon. The number of diagonals is equal to number of cables.
Formula for the number of diagonals: $\displaystyle \frac{n(n-3)}{2}$
where n is the number of sides (or telephones)

Last edited by sf7; April 3rd, 2014 at 06:11 AM.

April 3rd, 2014, 06:09 AM   #5
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 Originally Posted by eddybob123 Are you familiar with binomial coefficients? Since a cable connects two phones, and there are 10 of them, then there are 10C2 = 45 cables in total. With 20 phones, there are 20C2 cables in total.
eddybob123, could you explain in an example how it is calculated? Thanks in advance!

April 3rd, 2014, 04:41 PM   #6
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Quote:
 Originally Posted by eddybob123 You mean that there's cable connecting every single telephone? That is really inefficient... Are you familiar with binomial coefficients? Since a cable connects two phones, and there are 10 of them, then there are 10C2 = 45 cables in total. With 20 phones, there are 20C2 cables in total.
Thank you! Yes i am familiar with it, i used this method after i started drawing diagrams for this problem. Thank you!

April 4th, 2014, 11:23 AM   #7
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Quote:
 Originally Posted by sf7 The easiest way to solve this problem is finding the number of diagonals in polygon.
One needs to add to that the number of sides the polygon has. The resulting formula is the same as the formula for the binomial coeffiicient C(n, 2), where n is the number of sides.

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