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ldacosta1 · Oct 1, 2008, 07:11 PM

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?

Flying Blue Eagle · Oct 1, 2008, 09:08 PM

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.

FlyYakker · Oct 7, 2008, 08:17 PM

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.

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.

Wondergirl · Oct 7, 2008, 08:40 PM

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.)