What is a 'say now' number?

What are the number of ways in which the sum of 2 integers is less than or equal to a certain number, say N?

• How do I approach solving this problem? Please explain. Say N is 50. I asked a friend, what are the number of ways in which the sum of 2 POSITIVE integers is less than or equal to 50? He said the answer is 52C2, but he doesn't know how, he just knows the "formula". I again asked, what if I say natural numbers instead of positive integers? He said it would be 51C2. I thought of generalizing the question - and hence integers, not only POSITIVE integers.

• Answer:

  This cannot be generalized for Integers, since the number of ways will always be infinite. Now for your question, suppose you have to find the number of ways in which the sum of two positive numbers is less than or equal to 50. Then you could write this down as: [math]x+y \le 50[/math] with [math]x,y \geq 0[/math]. (this will change to [math]x,y \geq 1[/math] for natural numbers) And you could take an extra variable [math]t[/math] which takes the leftover value and always maintains the sum to 50 and re-write your equation as: [math]x+y+t=50[/math] with [math]x,y,t \geq 0[/math]. (this will again change to [math]x,y \geq 1[/math] for natural numbers) This could be solved easily using binomial expansion. Just find the coefficient of [math]x^{50}[/math] in the expansion of [math](1-x)^{-3}[/math] (I am assuming you'll know how we arrive at  [math](1-x)^{-3}[/math])   If you look closely you will notice the answer is only finite because you can limit the values of [math]x,y \geq 0[/math], once you remove that limitation (make them Integers), there will always be infinite ways to do it. Good Luck!