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Odds of Order

If 'A' is a single event...
and 'B' is a single event...
and 'C' is a single event...
They can be lined up in 6 different Orders, i.e:
1.  A,B,C
2.  B,C,A
3.  C,A,B
4.  A,C,B
5.  B,A,C
6.  C,B,A
Therefore there are 6 orders to putting these 'events' in.
When you have four events, A,B,C,D, you have maximum 24 orders to order these 'events' in.
What is the formula to working out ten 'events', i.e. A,B,C,D,E,F,G,H,I,J? 

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Sebin L. | Cornell Student SAT/ACT, K-12 English/Math/History/ScienceCornell Student SAT/ACT, K-12 English/Ma...
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Hi Thomas,
One way I personally use to visualize this type of problem is to map out ten positions/spots for the ten "events," and then write how many different events can fill each spot. So in the case of 10 events (A,B,C,D,...J), in the first spot, you can put any of the ten events in there. Then in the second spot, you can put any of the remaining events (9 of them left, right?). And if you continue in this manner, you'll reach the tenth spot, and when you go to place an event there, there will only be one possible event to add.
So the total number of possible permutations without repetition would be 10x9x8x7x6x5x4x3x2x1. That long chain of numbers is called a factorial - you can represent it with 10!. You can do this with other numbers: 3! = 3x2x1 and oddly enough, 0! = 1.
In short, the formula is 'n!' - n being the total number of events you need to find the permutations for the problem. This way, you don't have to write out each and every permutation. 
Joe B. | Math/Physics TutorMath/Physics Tutor
To work out the number of possibilities of ordering some number of distinct 'events', you simply take the factorial of the number of events you have. Factorial is that number multiplied by ever counting number less than it down to 1.
For example:
3! (read three factorial) = 3*2*1=6
4! = 4*3*2*1=24
10! = 10*9*8*7*6*5*4*3*2*1=3,628,800