2006-12-03 15:39:03 · 5 answers · asked by Starwars24 2 in Science & Mathematics ➔ Mathematics
(2n)!/(2n-3)!
= [ (2n)(2n-1)(2n-2)(2n-3)... 3*2*1] / [(2n-3)(2n-4)...3*2*1]
= 2n*(2n-1)*(2n-2)
or
= 4n*(2n-1)*(n-1)
2006-12-03 15:42:46 · answer #1 · answered by Scott R 6 · 1⤊ 0⤋
Let 2n = x
x!/(x-3)! = x*(x-1)*(x-2) = x(x^2-3x+2) = x^3 - 3x^2 + 2x
Now substitute 2n:
(2n)^3 - 3(2n)^2 + 2(2n) = 8n^3 - 24n^2 + 4n
2006-12-03 15:44:54 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Are people on crack?
use the identity
n! = n(n-1)!
= n(n-1)(n-2)!
= n(n-1)(n-2)(n-3)!
so: (2n)! = (2n)(2n-1)(2n-2)(2n-3)!
divide this by (2n-3)! and you get (2n)(2n-1)(2n-2)
2006-12-03 15:51:59 · answer #3 · answered by William M 2 · 1⤊ 1⤋
(2n)! = 2n (2n-1) (2n-2) [(2n-3)!]
So (2n)!/ (2n-3)! = 2n(2n-1)(2n-2)
2006-12-03 15:45:40 · answer #4 · answered by Jerry P 6 · 0⤊ 0⤋
2n*(2n-1)*(2n-2)=2n(4n^2-6n+2)=8n^3-12n^2+4n
2006-12-03 16:24:55 · answer #5 · answered by yupchagee 7 · 0⤊ 0⤋