Một số chuyên đề về tổ hợp dành cho học sinh có năng khiếu toán bậc trung học phổ thông

M é t s è c h u yª n đ ề v ề t ổ h ợp d n h ch o hä c si n h c ó n ă n g k hi ếu t o ¸ n bË c tr u n g h äc p h ỉ t h «n g 10 Ei (i = 1, , k) k (i) (ii) Ei ni k (n1 + n2 + + nk ) 18 12 18 + 12 = 30 E 10 F 10 E F E F Ei (i = 1, , k) n1 E2 k E1 n2 E1 E3 E1 4+4−1 = n3 E2 Ek k−1 nk k n1 n2 n3 nk 10 (i) 6.8.10 = 480 (ii) + + 10 = 24 38 = 6561 X n r n r n X r X n X X r n {1, 2, 3, 4, 5} r {2, 3, 4} P (n, r) {2, 4, 3} r X = X r n X C(n, r) n! (n − r)! n! P (n, r) = = C(n, n − r) (ii) C (n, r) = r! r!(n − r)! (i) P (n, r) = m! ≡ (1).(2) (m) (i) r 0! ≡ n (n − 1) X r (n − 1) n(n − 1) r P (n, r) = n(n − 1) (n − r + 1) = (ii) C(n, r) n! (n − r)! r r n X X r r P (n, r) = P (r, r) + P (r, r) + + P (r, r) r X C(n, r) P (n, r) = C(n, r)P (r, r) = C(n, r)r! r X (n − r) C (n, r) = C (n, n − r) n P (n, n) = n! r n r n n 12 10 11 r r 12 10 10 C(12, 4) = n 11 12 12! = 495 4!8! 12 C(10, 4) = 210 C(9, 3) = 84 11 10 495.210.84 = 8731800 n (n + 1) n=3 n+1=4 n kn + k k+1 1.3.1 n=3 k+1 = kn + = 16 16 20 r = 4, 5, 6, 7, 8, ∗) r = = k+1 k=3 n=3 kn + = 3.3 + = 10 ∗) r =5=k+1 k=4 r=5 n=2 kn + = 4.2 + = + = 13 ∗) r =6=k+1 k=5 + kn + = + 5.2 + = 15 ∗) r = = k+1 ∗) r = = k+1 k = + kn + = + 6.2 + = 17 k=7 + + kn + = + + 7.1 + = 19 ∗) r = = k+1 + + kn + = + + 8.1 + = 20 S x1 n vr x3 ≥ x2 vr x2 ≥ x1 xn ≥ xn−1 r    n(r − 1) + 1,       (n − 1)(r − 1) + + x1 ,    vr = (n − 2)(r − 1) + + x1 + x2 ,           (1)(r − 1) + + x + x + + x , n−1 r ≤ x1 x1 < r ≤ x2 x2 < r ≤ x3 xn−1 < r ≤ xn x x [x] x m n h (m − 1) i p= n p m −  = m−1 < m np ≤ n n p+1 p= h 25 i 26 m = 26 =3 X r n X r≤n X r=n X X = {A, A, B, B, B, C, C} ni (i = 1, 2, , k) r + nk = r ≤ n n AABCBBC k+2 P (n, r) P (n; n1 , n2 , , nk ) ≡ n1 !n2 ! nk ! n1 + n2 + 2; p2 = 3; p3 = 5; p4 = h 100 i h 100 i h 100 i h 100i S1 = + + + = 117 i h h 2100 i 3h 100 i5 h 100 100 i h 100 i h 100 i + + + + + = 45 S2 = (2).(3) (2).(5) (2).(7) (3).(5) (3).(7) (5).(7) i h 100 i h 100 i h 100 i h 100 + + + =6 S3 = (3).(5).(7) (2).(5).(7) (2).(3).(7) (2)(3).(5) h i 100 S4 = =0 (2).(3).(5).(7) π(100) = 100 − + − 117 + 45 − + = 25 30 15 3 A B C S1 = 15 + + = 19 S3 = X x = 30 − 29 + S2 − = S2 − n(A ∩ B ∩ C) =    n(A ∩ B) ≥    n(A ∩ C) ≥     n(B ∩ C) ≥ S2 ≥ x≥9−2=7 P X = {x1 , x2 , , xn } Dn P (xi ) 6= xi i = 1, 2, , n X = {x1 , , xn } h 1 1 Dn = n! − + − + (−1)n n! 1! 2! Q X n! Ai Q n(Q) = xi (i = 1, 2, , n) Sk = X n(Ai1 ∩ Ai2 ∩ ∩ Aik ) = C(n, k)(n − k)! = n! k! h 1 Dn = n(A1′ ∩ A′2 ∩ ∩ A′k ) = n! − + − + (−1)n n! 1! 2! Dn n = 1, 2, , 10 Dn n n n n! Dn n!Dn n a) b) a) Dn (Dn )2 Dn b) Ai i (i = 1, 2, , n) n(X) = (n!)2 ; Sr = C(n, r)[(n − r)!]2 , (r = 1, 2, , n) n(X) − S1 + S2 − + (−1)n Sn 8 8! − D8 = 40320 − 14833 = 25487 X n n0 ∈ X n0 f (n0 ) = n0 n! − Dn 6 a) b c) d) 6! 6! (6!)2 a) 6! D6 b) 6!D6 D6 = 6! (6!)2 6 c) 6!.1 = 6! (6!)2 6.(1).D5 D5 6!.(6).(1).D5 = 5! (6!)2 d) a) c) 1− D6 D5 − 5! 6! n×n n2 k (n − k) − k = n − 2k n−k n2 a b 4−x x c A = {1, 2, , 100} a+b = c+d e + f = 2g n ≥ 2) A = {0, 1, , 8} i+j =8 f (i) + f (j) = f : A −→ A S = {1, 2, , 280} P 1≤i1 < {1, 2, , n} A = {a1 , a2 , , an } ∈ N ∗ k∈Z + aj − k i, j ∈ {1, 2, , n} m n ∈ N∗ P 2k n + n+1 = n+1 C(n, k) k=0 k=1 k n P k ∈ [−n, n] | P (x) |≤ 22n P (x) ∈ R[x] | P (x) |≤ ≤ 2n x0 < x1 < < xn xn + a1 xn−1 + + an x ∈ [−n, n] P (x) = P (xj ), j = 0, 1, , n n! 2n F (x) ∈ Z(x) ≤ F (c) ≤ k c ∈ {0, 1, , k + 1} F (0) = F (1) = = F (k + 1) (1.x + 2.x2 + + n.xn )2 = a0 + a1 x + a2n x2n n(n + 1)(5n2 + 5n + 2) an+1 + an+2 + + a2n = 24 x ∈ N∗ P (x) ∈ Z[x] (bn ) d ∈ N∗ P (x) > x b1 = 1, bk+1 = P (bk ), ∀k ≥ (bn ) P (x) = x + 1, ∀x ∈ N ∗ p1 , p2 , , pn ∈ [0, 1] 1 1 ≤ 8n(1 + + + + ) | x − p | 2n −1 i i=0 n P n ∈ N∗ Sn = C(2n, n) n+1 Sn = [n/ P2] k=0 [0, 1] (C (n, k) − C (n, k − 1))2 α1 , α2 , , αn α1 , α2 , , αn c | α1 m1 + + αn mn |< | m1 | + + | mn |n−1 {1, 2, , 14} M = {1, 2, , 27} (m1 , m2 , , mn ) ∈ Z n A = {a1 , , ak } ⊂ q = 2p    a x + + a1q xq =    11     ap1 x1 + + apq xq = aij ∈ {−1, 0, 1} (0, 0, , 0), xj ∈ Z (x1 , , xq ) | xj |≤ q, ∀j = 1, 2, , q n+1 |T | = 49 S x∈M S = {1, 2, 3, , 100} P M S T ∈P M \ {x} S x 2n+2 − Zn = {0, 1, 2, , n − 1} An = {(a, b, c) : a, b, c ∈ Zn , a < b < c, a + b + c ≡ 0(modn)} Bn = {(a, b, c) : a, b, c ∈ Zn , a ≤ b ≤ c, a + b + c ≡ 0(modn)} |Bn | = |An | + n |An | |An+3| = |Bn | |An+3| = |An | + n T m |An | X = {1, 2, , 2000} 402 (2 + 22000) X (m, n) n (r, s), (r + 1, s + 1), (r + 2, s + 2) ≤ s ≤ 1990 T ≤ r ≤ 1989 (i, j) i j (i, j) An (i + 1, j ) (i, j + 1) Bn An > Bn n0 ≥ n0 n2 n1 n2 n1 ps n0 ≤ n1 ≤ n0 n2 n0 ∈ {2, 3, 4, 5} n0 ∈ {6, 7} k, n 3k n A1 A2 A2 000 3k 31250000 n n (a1 , a2 , , an ) a = n! 1, 2, , n S(a) = n X 2i i=1 b c b c n! S(b) − S(c) P1 , P2 , , Pn N, ≤ p ≤ n n n a1 = p, a2 = p(p − 1), an = (p − 1)an−2 + (p − 2)an−1 p p∈