Question 896212
Let's say we had x members total


That means we have x choices for president (assuming they are all candidates). 


That leaves us with x-1 choices for treasurer. We cannot pick the same person to run both positions.


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So we have x*(x-1) ways to pick 2 people (from a pool of x people)


We know that there are 506 ways to do this, so x*(x-1) = 506


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Solve for x


x*(x-1) = 506


x^2-x = 506


x^2-x-506 = 0


Use the quadratic formula to get the solutions x = 23 or x = -22. 


I show the steps in solving <a href="http://www.algebra.com/algebra/homework/quadratic/Quadratic_Equations.faq.question.896220.html">here</a>.


x is the number of people, so it has to be a positive whole number. That means we ignore x = -22


The only practical solution is x = 23.


That means the club has <font size=4 color="red">23 members</font>.


Notice how


x*(x-1) = 506


23*(23-1) = 506


23*22 = 506


506 = 506


So that confirms the answer.