word problem.... very tricky

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• December 19th 2012, 02:22 PM
asilvester635
word problem.... very tricky
can anyone explain how he got 28????
• December 19th 2012, 02:47 PM
skeeter
Re: word problem.... very tricky
count 'em ...

http://intermath.coe.uga.edu/tweb/CP...s/image001.gif

or think ...

2 computers ... 1 wire
3 computers ... 3 wires
4 computers ... 6 wires
5 computers ... 10 wires

see the pattern?
• December 19th 2012, 02:57 PM
Plato
Re: word problem.... very tricky
Quote:

Originally Posted by asilvester635
can anyone explain how he got 28????

This is known in graph theory as a complete graph of order eight: $\mathbb{K}_8$.

The number of edges in $\mathbb{K}_n$ is $\binom{n}{2}=\frac{n(n-1)}{2}.$
• December 19th 2012, 03:48 PM
cathectio
Re: word problem.... very tricky
Yes, the sum of numbers up to n is (n*(n-1))/2; therefore, for the n=8, the answer is 56/2 = 28. We need one fewer wire, since we are not connecting a computer to itself, as explained in the answer already provided. We need that for all 8 computers, considered as a pair each time. Therefore 8*7 / 2 is both logical and correct.
• December 19th 2012, 03:55 PM
Plato
Re: word problem.... very tricky
Quote:

Originally Posted by cathectio
Yes, the sum of numbers up to n is (n*(n-1))/2; therefore, for the n=8, the answer is 56/2 = 28. We need one fewer wire, since we are not connecting a computer to itself, as explained in the answer already provided. We need that for all 8 computers, considered as a pair each time. Therefore 8*7 / 2 is both logical and correct.

NO! That is absolutely not correct.
The formula I gave you in no way implies that a vertex is connected to itself.
This is such a well-known question, why do you question my reply?
• December 19th 2012, 06:07 PM
cathectio
Re: word problem.... very tricky
I reiterate what was posted: we need one fewer wire. I added that we are NOT connecting a computer to itself, as you seem to imply the opposite.
If one draws out the connections, they come to 28. BUT THE LOGICAL ANSWER GIVEN, THAT WE NEED ONE FEWER WIRE FOR ALL 8 CONNECTIONS RESULTS IN THE FORMULA I GAVE, N(N-1)/2.
What part of that do you have a problem with, Plato?