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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!
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TomasB Inner circle Sweden 1144 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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