f4 Binomial Expansion

nC4 + nC6
= n! / [4!(n-4)!] + n! / [6!(n-6)!]
抽出較小的為因式
= {n! / [4!(n-6)!]}{1/[(n-5)(n-4)] + 1/(5*6)}
= {n! / [4!(n-6)!]}[30 + (n-5)(n-4)] / [(n-5)(n-5)(5)(6)]
= {n! / [6!(n-4)!]}(30 + n^2 - 9n + 20)
= {n! / [6!(n-4)!]}(n^2 - 9n + 50)
(nC5)(2) = 2n! / [5!(n-5)!]
{n! / [6!(n-4)!]}(n^2 - 9n + 50) = 2n! / [5!(n-5)!]
約n!, 5!, (n-5)!
{1 / [6(n-4)]}(n^2 - 9n + 50) = 2
n^2 - 9n + 50 = (2)(6)(n-4)
n^2 - 9n + 50 = 12n - 48
n^2 - 21n + 98 = 0
(n - 7)(n - 14) = 0
n = 7 or n = 14

2009-12-15 18:29:27 補充：
第五行寫錯:= {n! / [4!(n-6)!]}[30 + (n-5)(n-4)] / [(n-5)(n-4)(5)(6)]

2009-12-15 20:20:24 補充：
{n! / [6!(n-4)!]}(n^2 - 9n + 50) = 2n! / [5!(n-5)!]
兩邊分子約去n!
左邊分母6! = 6*5*4*3*2*1,右邊分母5!=5*4*3*2*1,兩邊約去5*4*3*2*1左邊餘下6
左邊分母(n-4)! = (n-4)(n-5)(n-6)...(3)(2)(1),右邊分母(n-5)! = (n-5)(n-6)...(3)(2)(1)
兩邊約去(n-5)(n-6)...(3)(2)(1),左邊餘下(n-4)