Counting Discrete Math

Question

Consider the equation a + b + c = 202. How many solutions are there where a,b, and c are all nonnegative integers?

Answer 1

Nonnegative integers means 0, 1, 2, 3, . . . 202.

Start with a = 0

Then b could be any number from 0 to 202 or 203 possibilities

Which means that c would be defined. Once you pick "b" then "c" would be only one possible number to get the sum to be 202.

So with a = 0, there are 203 possible solutions.

Now, let a = 1

Then, "b" could be any number from 0 to 201 but it can no longer be 202, There are consequently 202 possible solutions (again, when you pick a = 1 and pick "b" at whatever, then "c" is defined as one specific number that would yield a sum of 202.

Now let a = 2

This time "b" must be between 0 and 200 so there are 201 possible solutions.

Skip to a = 200. Then b could be 0, 1, or 2. Again "c" would be defined once "b" is picked so there are 3 possible solutions.

For a = 201. There are only two possibilities, b = 1 with c = 0 or b = 0 with c = 1.

When a = 202, there is only one possibility, b = 0 and c = 0.

So can you see that the total number of possible solutions is:

203 + 202 + 201 . . . + 103 + 102 + 101. . . + 3 + 2 + 1?

To determine what this total is, think of the numbers as pairs:

(203 + 1) + (202 + 2) + (201 + 3) . . . + (103 + 101) + 102

or 204 + 204 + 204 . . . + 204 + 102

or 101 pairs (each adding to 204) plus a stand-alone = 204•101 + 102 = 20706