# 09 análise combinatória - parte ii (fatorial)

11 de Jun de 2012
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### 09 análise combinatória - parte ii (fatorial)

• 1. Análise Combinatória – Parte II Soraya Mara Menezes de Souza
• 2. Análise Combinatória – Parte II Fatorial Sendo n um número inteiro maior que 1, define-se fatorial de n como o produto dos n números naturais consecutivos de n a 1. n! = n ( n − 1) ( n − 2) ( n − 3) ... 3 ⋅ 2 ⋅ 1 sendo n ∈ Ν e n > 1 6! =6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 6! = 6 ⋅ 5! 5! =5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 6! = 6 ⋅ 5 ⋅ 4! 7! =7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1
• 3. Análise Combinatória – Parte II Fatorial Calcule: 8! 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1 40.320 = = = 336 5! 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1 120 ou 8! 8 ⋅ 7 ⋅ 6 ⋅ 5! 8 ⋅7 ⋅6 = = = 336 5! 5! 1
• 4. Análise Combinatória – Parte II Fatorial Calcule: 8! 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4! 1680 = = = 840 4! ⋅ 2! 4! ⋅ 2 ⋅ 1 2 5! ⋅ 4! 5! ⋅ 4 ⋅ 3! 4 2 = = = 3! ⋅ 6! 3! ⋅ 6 ⋅ 5! 6 3
• 5. Análise Combinatória – Parte II Fatorial Resolva as equações: x! = 15 ( x − 1)! ( n − 2)! = 720 x ( x − 1)! = 15 ( x − 1)! ( n − 2)! = 6! 15 ( x − 1)! x = n −2 = 6 ( x − 1)! x = 15 n =8
• 6. Análise Combinatória – Parte II Fatorial Resolva as equações: x! ( n − 2)! = 2 ( n − 4)! = 30 ( x − 2)! ( n − 2)(n − 3)(n − 4)! = 2( n − 4)! x ( x − 1)( x − 2)! ( n − 2)(n − 3) = 2( n − 4)! = 30 ( n − 4)! ( x − 2)! ( n − 2)(n − 3) = 2 x ( x − 1) − 30 = 0 n ² − 5n + 6 = 2 x ² − x − 30 = 0 n ² − 5n + 4 = 0 x = 6 ou x = −5 n = 4 ou n = 1 x =6 n =4
• 7. Análise Combinatória – Parte II Fatorial Simplifique as expressões: n! n! − ( n + 1)! ( n − 1)! n! n( n − 1)! n! − (n + 1)n! n! (1 − n − 1) =n = = −n ( n − 1)! n! n! ( n + 2)! ( n + 2)(n + 1) n( n − 1)! n(n + 2)(n + 1)   ( n − 1)! ( n − 1)!
• 8. Análise Combinatória – Parte II Soraya Mara Menezes de Souza