彰師大105年度離散數學問題10?

(a)
非負整數解數
= H( 4 , 12 )
= C( 4+12-1 , 12 )
= C( 15 , 12 )
= C( 15 , 3 )
= 15*14*13 / (3*2*1)
= 455 ..... Ans

(b)
非負整數解數
= Σ H( 4 , k ) , from k = 0 to k = 11
= Σ C( 4+k-1 , k )
= Σ C( k+3 , k )
= Σ C( k+3 , 3 )
= Σ (k+3)(k+2)(k+1)/6
= (1/6) * Σ (k+3)(k+2)(k+1)
= (1/6) * Σ ( k³ + 6k² + 11k + 6 ) , from k = 0 to k = 11

當 k = 0 , ( k³ + 6k² + 11k + 6 ) = 6

Σ ( k³ + 6k² + 11k + 6 ) , from k = 1 to k = 11
= Σk³ + 6Σk² + 11Σk + Σ6
= n²(n+1)²/4 + n(n+1)(2n+1) + 11n(n+1)/2 + 6n , 其中 n = 11
= 11²12²/4 + 11*12*23 + 11*11*12/2 + 6*11
= 4356 + 3036 + 726 + 66
= 8184

非負整數解數
= (1/6) * Σ ( k³ + 6k² + 11k + 6 ) , from k = 0 to k = 11
= (1/6) * ( 6 + 8184 )
= 1365 ..... Ans