Suppose several people enter a drawing to win one of two prizes. If the prizes are identical and each person is allowed only one entry in the drawing, how many different winning pairs are possible?
Two entries
One winning pair
Three entries
Three winning pairs
Four entries
Six winning pairs
Five entries
Ten winning pairs
To find the number of winners, one thing we can do is line all n people up, and because an individual can't win both prizes, think about matching each of them with the other n – 1 people. This gives us a total of n(n – 1) winning pairs. However, each match appears twice because we lined up all the people. So, we have to divide by 2 in order to arrive at the total number of unique winning pairs. Notice that this is exactly how we counted the number of airways between airports in our first solution to the Flight Paths problem!
Another way to look at our prize winning pairs is in terms of combinations. Remember, that permutations give us all combinations where order matters. Since we don't distinguish between person 1 winning with person 2 and person 2 winning with person 1, we can calculate the number of prizes by finding the number of combinations of 2 people selected from a group of n people. Mathematically, we refer to this as n choose 2 and we can represent this visually as in the diagram above.
In the example above we have four people and we want to look at all of the possible pairs. We can pair the black with blue, black with purple, and black with green, and so on. Each of these possible pairs is represented by the dual-colored circle where you can imagine diagonal lines through each individual would intersect. Do you recognize our triangular numbers? Our expression for the solution is adjusted by 1, of course, since our first combination starts with two people instead of one, as in our Flight Paths problem. This gives us yet another way to express the solution to our Triangular Numbers problem using combinations.