

10cardsdown Special user Out There Somewhere 664 Posts 
If there were 9 different numbers, say the numbers 1 through 9, and you asked someone to arrange these 9 numbers at random, how many possibilities are available and how would the end result be calculated? Thanks!

TomasB Inner circle Sweden 1143 Posts 
You can choose the first digit as any of the 9. So 9 possibilities for the first digit. Once that digit is placed in position you have 8 digits left to place in the second position. Once that digit is chosen you have 7 left to choose from, and so on until you are left with a single digit which you have to place in the last position. The number of ways to arrange 9 different digits is thus 9*8*7*6*5*4*3*2*1 = 362880, or more compactly written as 9! where ! means "factorial".
/Tomas 
10cardsdown Special user Out There Somewhere 664 Posts 
TomasB,
Thanks so much, you always figure these things out for me. Thanks so much, greatly appreciated! 
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