Probabilités 1 Analyse combinatoire

m
n mn
(1,1),(1,2),··· ,(1, n)
(2,1),(2,2),··· ,(2, n)
(m, 1),(m, 2),··· ,(m, n)
r
n1
n2
n1n2···nrr
3
4 5
2 4

3×4×5×2 = 120
7
3 4
263×104= 175 760 000
26 ×24 ×10 ×9×8×7 = 78 624 000
E n E
{0,1}2n
a, b, c
abc acb bac bca cab cba
E n
E n E
n
n! = n(n−1)(n−2) ···3.2.1
P EP P ER
P E 6!
P1P2E1P3E2R P1P3E2P2E1R
P E 6!
3!2! = 60
E n p 1≤p≤n
p E p E
p E
Ap
n=n(n−1)(n−2) ···(n−p+ 1) = n!
(n−p)!

n p 0≤p≤n E
n p E p
p E n
Cp
n=n
p=n!
p!(n−p)!
p > 0p E
p p E
Ap
n=Cp n×p!
p= 0 E0∅
A0
n=C0 n
(a+b)n=C0
nan+C1 nan−1b+··· +Cp nan−pbp+··· +Cn nbn
(a+b)n= (a+b)(a+b)···(a+b)
a p b n −p
Cp
n=Cn−p n
Cp
n=Cp−1
n−1+Cp n−1
Cp
n=Cp−1
n−1+Cp−1 n−2+··· +Cp−1 p−1
Cp
n=n
pCp−1
n−1
2n=C0
n+C1 n+··· +Cp
n+··· +Cn
n
(1 + 1)n

E
E
{1,2,3,4,5,6}
{g, f}
7!
Ω
Ω
Ω
AΩ
Ω∈ A
A∈ A ⇒ CΩA∈ A
Ai∈ A, i ∈ ⇒ ∪i∈Ai∈ A
∅ ∈ A
∩i∈Ai‘∈ A ∩Ai=CΩ(∪(CΩAi))
A
Ω∅
A=CΩA A
A B A ∩B=∅
(Ω,A) Ω
ΩA={Ω,∅}
A={Ω, A, A, ∅}
A=P(Ω)
Ω
A
∅
(Ω,A)P:A →
[0,1]