# 點算的奧秘：帶上、下限的排列組合問題

|S1 ∪ S2 ∪ ... Sn| = Σ1 ≤ k ≤ n (−1)k − 1 Sn, k     (1)
|S1' ∩ S2' ∩ ... Sn'| = |U| + Σ1 ≤ k ≤ n(−1)k Sn, k     (2)

|S1 ∪ S2 ∪ ... Sn| = Σ1 ≤ k ≤ n (−1)k − 1 C(n, k) nk     (3)
|S1' ∩ S2' ∩ ... Sn'| = |U| + Σ1 ≤ k ≤ n(−1)k C(n, k) nk     (4)

|U| − S3, 1 + S3, 2 − S3, 3 = 21 − 20 + 3 − 0 = 4 □

Sn, 1 = Σ |Si| = Σ C(n + r − m − di − 2, n − 1)

Sn, 2 = Σ |Si ∩ Sj| = Σ C(n + r − m − di − dj − 3, n − 1)

Sn, k = Σ |Sx ∩ ... Sz| = Σ C(n + r − m − dx − ... dz − k − 1, n − 1)

 C(n + r − m − 1, n − 1) + Σ1 ≤ k ≤ n [(−1)k Σ C(n + r − m − dx − ... dz − k − 1, n − 1)]     (5) 其中m = m1 + ... mn；di = Mi − mi (1 ≤ i ≤ n)

 |U| + Σ1 ≤ k ≤ n(−1)k C(n, k) nk = C(n + r − nm − 1, n − 1) + Σ1 ≤ k ≤ n(−1)k C(n, k) C(n + r − nm − k(M − m + 1) − 1, n − 1) = Σ0 ≤ k ≤ n(−1)k C(n, k) C(n + r − nm − k(M − m + 1) − 1, n − 1)     (6)

 C(3, 0) × C(9, 2) − C(3, 1) × C(4, 2) + C(3, 2) × C(−1, 2) − C(3, 3) × C(−6 , 3) = 1 × 36 − 3 × 6 + 3 × 0 − 1 × 0 = 18 □

|U| − S2, 1 + S2, 2 = 34 − (1 + 24) + 1 = 65 □

 r! Σ ---------------------- r1! × r2! × ... rn!

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