設 0≦m≦10、0≦n≦10，若 m 顆紅球和最多 n 顆黑球(即 0 或 1 或 2 或……或 n 顆黑球) ，排成一列的方法數共有C10取4種，試求數對( m , n )＝? ANS(3,6) 或(5,4)?

(m+0)!/(m! 0!) + (m+1)!/(m! 1!) + (m+2)!/(m! 2!) + ... + (m+n)!/(m! n!) = 10C4

m! + (m+1)!/1! + (m+2)!/2! + ... + (m+n)!/n! = 210 m!

1*2*3*...*m + 2*3*4*...*(m+1) + 3*4*5*...*(m+2) + ... + (n+1)(n+2)(n+3)*...*(m+n) = 210 m!

1*2*3*...*(m+1) + 2*3*4*...*(m+2) + 3*4*5*...*(m+3) + ... + (n+1)(n+2)(n+3)*...*(m+n+1)
- ( 0*1*2*...*m + 1*2*3*...*(m+1) + 2*3*4*...*(m+2) + ... + n (n+1) (n+2) * ... * (m+n) ) = 210 m! (m+1)

(n+1)(n+2)(n+3)*...*(m+n+1) = 210 (m+1)!
(n+1)(n+2)(n+3)*...*(m+n+1) = 5*6*7 (m+1)!
若 m ≥ 6 , 則 5*6*7 (m+1)! 必被 7² 整除, 那麼 (n+1)(n+2)(n+3)*...*(m+n+1) 中必包含 7*8*9*10*11*12*13*14 ,
但 (m+1)! 不被13整除,矛盾! 故 m ≤ 5.
當 m = 5 , 5*6*7*6! = 5*6*7*8*9*10 , n = 5-1 = 4
當 m = 3 , 5*6*7*4! = 7*8*9*10 , n = 7-1 = 6