Toan-daisotohop-chuong2 - AI SO TO HP Chng II HOAN V 1....

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ÑAÏI SOÁ TOÅ HÔÏP Chöông II HOAÙN VÒ 1. Giai thöøa Vôùi soá nguyeân döông n, ta ñònh nghóa n giai thöøa, kí hieäu n!, laø tích caùc soá nguyeân lieân tieáp töø 1 ñeán n. n! = 1.2.3…(n – 2) (n – 1)n. Vì tieän lôïi, ngöôøi ta qui öôùc : 0! = 1. Töø ñònh nghóa, ta coù : n(n – 1) … (n – r + 1) = n! (n r)! vaø (n – 1)!n = n! Ví duï : a) 5! = 1.2.3.4.5 = 120; b) 9! 5! = 9.8.7.6 = 3024; c) 3!4 = 4! = 1.2.3.4 = 24; d) (n 2)! (n 3)! + = (n + 2)(n + 1)n(n – 1)(n – 2). 2. Hoaùn vò Coù n vaät khaùc nhau, saép vaøo n choã khaùc nhau. Moãi caùch saép ñöôïc goïi laø 1 hoaùn vò cuûa n phaàn töû. Theo qui taéc nhaân, choã thöù nhaát coù n caùch saép (do coù n vaät), choã thöù nhì coù n – 1 caùch saép (do coøn n – 1 vaät), choã thöù ba coù n – 2 caùch saép (do coøn n – 2 vaät), …, choã thöù n coù 1 caùch saép (do coøn 1 vaät). Vaäy, soá hoaùn vò cuûa n phaàn töû, kí hieäu P n , laø : P n = n(n – 1)(n – 2)… × 1 = n! Ví duï 1. Töø 3 chöõ soá 1, 2, 3 coù theå taïo ñöôïc bao nhieâu soá goàm 3 chöõ soá khaùc nhau ? Giaûi
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Moãi soá goàm 3 chöõ soá khaùc nhau taïo ra töø 1, 2, 3 laø moät hoaùn vò cuûa 3 phaàn töû. Vaäy coù : P 3 = 3! = 6 soá. (caùc soá ñoù laø : 123, 132, 213, 231, 312, 321) Ví duï 2. Trong moät lôùp hoïc, thaày giaùo phaùt phieáu thaêm doø yeâu caàu hoïc sinh ghi thöù töï 3 moân Toaùn, Lyù, Hoùa ñang hoïc theo möùc ñoä yeâu thích giaûm daàn. Hoûi coù bao nhieâu caùch ghi khaùc nhau ? Giaûi Ñaây laø hoaùn vò cuûa 3 phaàn töû. Vaäy coù: P 3 = 3! = 6 caùch, khi ñoù coù 6 caùch ghi laø: (T,L,H), (T,H,L), (L,T,H), (L,H,T), (H,T,L), (H,L,T). Ví duï 3. Coù 2 saùch toaùn khaùc nhau, 3 saùch lyù khaùc nhau vaø 4 saùch hoùa khaùc nhau. Caàn saép xeáp caùc saùch thaønh moät haøng sao cho caùc saùch cuøng moân ñöùng keá nhau. Hoûi coù bao nhieâu caùch saép ? Giaûi Tröôùc tieân, ta saép theo moân thì coù P 3 = 3! = 6 caùch. Tieáp ñeán, caùc saùch töøng moân ñoåi choã cho nhau, toaùn coù P 2 = 2! = 2 caùch, lyù coù P 3 = 3! = 6 caùch, hoùa coù P 4 = 4! = 24 caùch. Vaäy, theo qui taéc nhaân, coù : 6 × 2 × 6 × 24 = 1728 caùch. Baøi 18. Giaûi phöông trình : x! (x 1)! (x −− + = 1 6 vôùi x ¥ * Giaûi + = 1 6 6[x! – (x – 1)!] = (x + 1)! 6[x(x – 1)! – (x – 1)!] = (x + 1)! 6(x – 1)!(x – 1) = (x + 1)x(x – 1)! 6(x – 1) = x(x + 1) x 2 – 5x + 6 = 0 x2 x3 = = Nhaän do x ¥ *. Baøi 19. Giaûi baát phöông trình : n4 nn 2 P P.P + + < n1 15 P (*) Ñieàu kieän n > 1, n . ¥
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Ta coù : (*) (n 4)! n!(n 2)! + + < 15 (n 1)! (n 4)(n 3)(n 2)! n(n 1)!(n + ++ −+ < 15 (n 1)! (n 4)(n 3) n + + < 15 n 2 + 7n + 12 < 15n n 2 – 8n + 12 < 0 2 < n < 6 Do ñieàu kieän neân n { } 3, 4, 5 . Baøi 20. Goïi P n laø soá hoaùn vò cuûa n phaàn töû. Chöùng minh : a) P n – P n-1 = (n – 1)P n-1 b) 1 + P 1 + 2P 2 + 3P 3 + … + (n – 1)P n-1 = P n .
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This note was uploaded on 11/26/2011 for the course MATH 1002 taught by Professor Chuck during the Spring '11 term at University of Western States.

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Toan-daisotohop-chuong2 - AI SO TO HP Chng II HOAN V 1....

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