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 The Magic Cafe Forum Index » » Magical equations » » Calculating 9 numbers (0 Likes)

 10cardsdown Special user Out There Somewhere 664 Posts Posted: Sep 20, 2005 10:09 pm    0 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 Posted: Sep 21, 2005 01:38 am    0 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 Posted: Sep 21, 2005 08:48 am    0 TomasB, Thanks so much, you always figure these things out for me. Thanks so much, greatly appreciated! The Magic Cafe Forum Index » » Magical equations » » Calculating 9 numbers (0 Likes)
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