How would you answer the following math problem:

You and 9 other people take turns riding in pairs on amusement park rides. How many different pairs are possible?

Idacosta1 - Stop to think ,this is simple,
You have a total of 9 people
how many pairs do you have rideing the rides, Uneven number of riders but I figure the odd one will ride with someone that makes another pair
(9 divided by 2 = 4- with one person left over, someone rides with him or herso you have one more pair to ride, which makes the answer ( 5 - pairs ) good luck and have a good day ;; GOD BLESS ::F.B.E.

Sorry Flying Blue Eagle but it's 10 people ( "you and 9 other"). Also this looks like a "combinations" question... that is, not just how many pairs, but HOW MANY DIFFERENT PAIRS. It is a larger number that you give. Look at Permutations and combinations (2) - Topics in precalculus (http://www.themathpage.com/aprecalc/permutations-combinations-2.htm).

I don't remember these types of questions when I was was a lad, but then there are a lot of things I don't remember any more.

My math teachers always told us, "Draw a picture."

10 people = A B C D E F G H I J

pairs possible = AB AC AD AE AF AG AH AI AJ, BC BD BE BF BG BH BI BJ, CD CE CF CG CH CI CJ, etc. You do the rest. (Remember, AB is the same pair as BA.)