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