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September 6th, 2017, 11:13 AM   #1
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(People and Relationships) Can You Find The Equation?

I discovered something cool about points and lines, wondering anyone else knew about this.

I call it people and relationships, point=people and line=relationship
There is 1 relationship between every 2 people.

Here is the table:
# people - # relationship
0-0
1-0
2-1
3-3
4-6
5-10


Take a look at this picture (Click)

There is an obvious pattern in the table. See if you spot it.
And see if you can find out the whole equation.
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September 6th, 2017, 11:48 AM   #2
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The sum of the numbers in each row is the second number in the next row. Is that what you mean?

The "whole equation" for what in terms of what?
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September 6th, 2017, 12:50 PM   #3
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Quote:
Originally Posted by skipjack View Post
The sum of the numbers in each row is the second number in the next row. Is that what you mean?

The "whole equation" for what in terms of what?
Yeah you spotted the pattern. But can you make up an equation for this pattern?

# people = x
# relationship = y

Example equation
y=4(x)²+5

Last edited by Logix; September 6th, 2017 at 01:22 PM.
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September 6th, 2017, 02:44 PM   #4
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$\displaystyle y = \binom{x}{2}$
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September 6th, 2017, 03:02 PM   #5
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Check your equations by substituting the numbers on my table into your equation. Hint: if you graph the table, it would look like a quadratic equation.

Last edited by Logix; September 6th, 2017 at 03:04 PM.
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September 6th, 2017, 03:57 PM   #6
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For example, $\displaystyle \binom52 = 5(4)/2 = 10$.
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September 6th, 2017, 04:23 PM   #7
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Quote:
Originally Posted by skipjack View Post
For example, $\displaystyle \binom52 = 5(4)/2 = 10$.
5 on top of 2, I have no idea about that. But well, you found the equation.
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September 6th, 2017, 05:32 PM   #8
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Quote:
Originally Posted by Logix View Post
5 on top of 2, I have no idea about that. But well, you found the equation.
That is combination of 2 people out of 5 people i.e. number of ways 2 people can be combined when we have 5 people to choose from.

$\displaystyle \binom {n}{x} = \frac {n!}{(n-x)!x!} $
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