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 2018-04-10 13:18:05 +0200 received badge ● Student (source) 2017-05-09 14:01:17 +0200 received badge ● Famous Question (source) 2016-02-08 17:09:50 +0200 received badge ● Taxonomist 2014-02-15 10:52:17 +0200 received badge ● Notable Question (source) 2013-01-31 08:33:32 +0200 received badge ● Popular Question (source) 2012-03-08 07:33:29 +0200 answered a question A proper combinatorics tutorial? I'm not sure if this is still terribly relevant for you any more but I have made a good start on a tutorial. http://sagenb.org/home/pub/4474/ I hope this is helpful. 2012-03-08 07:26:26 +0200 received badge ● Scholar (source) 2012-03-08 07:26:26 +0200 marked best answer calculating n!/(n-k)! for k-permutations Sure, what you are looking for is the falling factorial. for n in (0..7) : [falling_factorial(n,k) for k in (0..n)] [1] [1, 1] [1, 2, 2] [1, 3, 6, 6] [1, 4, 12, 24, 24] [1, 5, 20, 60, 120, 120] [1, 6, 30, 120, 360, 720, 720] [1, 7, 42, 210, 840, 2520, 5040, 5040]  2012-03-03 17:22:16 +0200 answered a question calculating n!/(n-k)! for k-permutations Thanks, I thought doing it this way might be really slow but after some testing it appears that even with large inputs it doesn't take long to calculate. 2012-03-03 17:15:23 +0200 received badge ● Supporter (source) 2012-03-03 17:01:08 +0200 answered a question calculating n!/(n-k)! for k-permutations I am not looking for the binomial coefficient, what the binomial(n,k) calculates is n!/(k!(n-k)!). What I am looking for is a the number of ways to get k-permutations out of n elements. Sorry if I was not clear. 2012-03-03 16:40:33 +0200 asked a question calculating n!/(n-k)! for k-permutations I was wondering if there is a way to calculate n!/(n-k)! in sage without using the factorial(n) method for n and n-k and then dividing.