Question 1083149
Let's compute the following ten nPr values. 
Fix n = 9 (ie don't change this value)
Let r be a whole number and let it range from r = 0 to r = 9.
The goal is to find which r value yields n P r = 9 P r = 3024.


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When r = 0, then...


n P r = (n!)/((n-r)!)
9 P 0 (9!)/((9-0)!)
9 P 0 = (9!)/(9!)
9 P 0 = (9*8*7*6*5*4*3*2*1)/(9*8*7*6*5*4*3*2*1)
9 P 0 = (362880)/(362880)
9 P 0 = 1


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When r = 1, then...


n P r = (n!)/((n-r)!)
9 P 1 (9!)/((9-1)!)
9 P 1 = (9!)/(8!)
9 P 1 = (9*8*7*6*5*4*3*2*1)/(8*7*6*5*4*3*2*1)
9 P 1 = (362880)/(40320)
9 P 1 = 9


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When r = 2, then...


n P r = (n!)/((n-r)!)
9 P 2 (9!)/((9-2)!)
9 P 2 = (9!)/(7!)
9 P 2 = (9*8*7*6*5*4*3*2*1)/(7*6*5*4*3*2*1)
9 P 2 = (362880)/(5040)
9 P 2 = 72


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When r = 3, then...


n P r = (n!)/((n-r)!)
9 P 3 (9!)/((9-3)!)
9 P 3 = (9!)/(6!)
9 P 3 = (9*8*7*6*5*4*3*2*1)/(6*5*4*3*2*1)
9 P 3 = (362880)/(720)
9 P 3 = 504


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When r = 4, then...


n P r = (n!)/((n-r)!)
9 P 4 (9!)/((9-4)!)
9 P 4 = (9!)/(5!)
9 P 4 = (9*8*7*6*5*4*3*2*1)/(5*4*3*2*1)
9 P 4 = (362880)/(120)
9 P 4 = 3024


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When r = 5, then...


n P r = (n!)/((n-r)!)
9 P 5 (9!)/((9-5)!)
9 P 5 = (9!)/(4!)
9 P 5 = (9*8*7*6*5*4*3*2*1)/(4*3*2*1)
9 P 5 = (362880)/(24)
9 P 5 = 15120


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When r = 6, then...


n P r = (n!)/((n-r)!)
9 P 6 (9!)/((9-6)!)
9 P 6 = (9!)/(3!)
9 P 6 = (9*8*7*6*5*4*3*2*1)/(3*2*1)
9 P 6 = (362880)/(6)
9 P 6 = 60480


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When r = 7, then...


n P r = (n!)/((n-r)!)
9 P 7 (9!)/((9-7)!)
9 P 7 = (9!)/(2!)
9 P 7 = (9*8*7*6*5*4*3*2*1)/(2*1)
9 P 7 = (362880)/(2)
9 P 7 = 181440


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When r = 8, then...


n P r = (n!)/((n-r)!)
9 P 8 (9!)/((9-8)!)
9 P 8 = (9!)/(1!)
9 P 8 = (9*8*7*6*5*4*3*2*1)/(1)
9 P 8 = (362880)/(1)
9 P 8 = 362880


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When r = 9, then...


n P r = (n!)/((n-r)!)
9 P 9 (9!)/((9-9)!)
9 P 9 = (9!)/(0!)
9 P 9 = (9*8*7*6*5*4*3*2*1)/(1)
9 P 9 = (362880)/(1)
9 P 9 = 362880


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In the work shown above, we see that 9 P 4 = 3024


So if 9 P r = 3024, then r = 4


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Answer: <font color=red>4</font>