ANÁLISE COMBINATORIA

Matemática

Prof.: Joaquim Rodrigues

ANÁLISE COMBINATÓRIA
FATORIAL: Sendo n um número natural maior que 1, definimos como fatorial de n e representamos por n! o número: n! = n ⋅ (n − 1) ⋅ (n − 2) ⋅ (n − 3) ⋅ ... ⋅ 3 ⋅ 2 ⋅ 1
Adotamos as seguintes definições especiais:
0! = 1
1!= 1
TABELA DE FATORIAIS DE 0 a 10
0! = 1
1!= 1
2! = 2 ⋅ 1 = 2
3! = 3 ⋅ 2 ⋅ 1 = 6
4 ! = 4 ⋅ 3 ⋅ 2 ⋅ 1 = 24
5 ! = 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 120
6 ! = 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 720
7 ! = 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 5040
8 ! = 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 40.320
9 ! = 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 362.880
10 ! = 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 3.628.800
Nota: A notação de fatorial afeta apenas o primeiro número antes do sinal de exclamação. Por exemplo:
2 ⋅ 3 ! = 2 ⋅ (3 ⋅ 2 ⋅ 1) = 2 ⋅ 6 = 12
(2 ⋅ 3)! = 6 ! = 720
QUESTÕES
Questão 01
Calcule:
5! + 3!
a)
2!
10 !
b)
8!
4 ! − 2! − 0 !
c)
1!
6 ! + 3 ! − 2!
d)
5!
e) 2 ⋅ 4 ! + 3 ⋅ 0 !
f) (2! ) 2 ⋅ (6 − 1)!
7! ⋅ 9!
g)
6! ⋅ 8!
12!
h)
8! ⋅ 5!

R: 63
R: 90
R: 21

181
30
R: 51
R:

R: 480

Questão 02
Simplifique as expressões: n! a)
R: n
(n − 1)!
(2n + 2)!
b)
R: 2n + 2
(2n + 1)!
(2n + 2)!
c)
R: (2n + 2)(2n + 1)
(2n)!
n! − (n + 1)!
d)
R: −n n! (n + 2)!
e)
R: n (n + 1)(n + 2)
(n − 1)!
(n + 5)!
f)
R: (n + 5)(n + 4)
(n + 3)!
(n + 2)! − (n + 1)!
g)
R: (n + 1) 2 n! Questão 03
Resolver a equação:
a) x ! = 15 ⋅ ( x − 1)!
b) (n − 2)! = 2 ⋅ (n − 4)! x! c)
= 30
( x − 2) ! m! + (m − 1)! 5
d)
=
(m + 1)! − m! 16
(n + 1)! − n!
e)
= 8n
(n − 1)!
Questão 04
Resolver a equação:
a) n! = 6
b) (n − 2)! = 1
c) (n − 9)! = 1
d) (n + 1)! = 24
e) (n − 4)! = 120
f) (n − 2)! = 720

R: 15
R: 4
R: 6
R: 4
R: 8

R: 3
R: 2, 3
R: 9, 10
R: 3
R: 5
R: 8

Questão 05
Sabendo que a n =
a) a10

R: 9

n! ⋅ (n 2 − 1)
, calcule:
(n + 1)!
b) a1980
R: 1979

R: 63
R: 99

Questão 06
Com quantos zeros termina o número 69!?
R: 13

1

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