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n = 11, r = 6
Good question. You usually do it the opposite way: given n and r, find nPr.
The formula is nPr = n!/(n-r)!. For example, if n is 6 and r is 2, then 6P2 = 6!/(6-2)! = 6!/4! =6x5x4x3x2x1/4X3x2x1.
Notice that (4x3x2x1) cancels, so that 6P2 = 6x5. Also, 5 = (6 - 2) +1.
So 332640 = n!/(n-r+1). We still can't find n and r without a second equation, so we have to resort to trial and error. If we factor 332,640, that may give us a hint:
332,640 = 2x166,320 = 2x(2x83,160) = 2^2x(2x41,580) = 2^3x(2x20,790) = 2^4x(2x10395) = 2^5x(5x2079) = 2^5x5x(3x693) = 2^5x3x5x(3x231) = 2^5x3^2x5x(3x77) = 2^5x3^3x5x(7x11).
This tells us that 11 is definitely in there. 7, or more likely 14 is in there. 5 tells us that either 10 or 15 is in there. 12^5 is 248832 so we guess 11x12x13x14x15 = 360,360 but ... whoops! there is no 13 in the factors, so that can't be right.
Maybe 12 is the highest number that can be in the answer: 12x11x10x9x8x7=665280. Nope.
How about 11x10x9x8x7x6 = 33,2640. Yes!!!
n=11, 11-r+1=6 -> 12 - 6 = r -> r=6.
Check: 11P6 = 11!/(11-6)! = 11!/5! = 11x10x9x8x7x6x5.../5x4... = 11x10x9x8x7x6 = 110x72x42 = 7,920x42 = 332,640.
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The number of permutations of n distinct objects taken r at a time is nPr equals n divided by n-r?
nPr is n!/(n-r)!. The ! is factorial; for example 5! = 5*4*3*2*1.
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