Corrigé de l`exercice 16 de la feuille de TD no 4

n p n >p>0
n
X
k=pk
p=n+ 1 p+ 1
E n + 1 E={1, . . . , n, n + 1}
A p + 1 E m
p+ 1 n+ 1
A m m 6p
A{1, . . . , p}p
card(A) = p+ 1 > p m >p+ 1 p+ 1
m+ 1
m∈ {p+1, . . . , n+1}A⊂E p+1
m
a1< a2<· · · < ap< ap+1 p+ 1 A A
ap+1 m p
a1, a2, . . . , ap{1, . . . , m −1}
f A ⊂E p + 1
m A0⊂ {1, . . . , m −1}
p f(A) = Ar{m}g
g(A0) = A0∪ {m}A0={a1, a2, . . . , ap}m−1
p
P:= Pp+1(E)p+ 1 E
card(P) = n+1
p+1m∈ {p+ 1, . . . , n + 1}Pm
A m
card(Pm) = m−1
pA
m∈ {p+ 1, . . . , n + 1}Pp+1, . . . , Pn+1 P
P
PmP=Pp+1 q · · · q Pn+1
n+ 1
p+ 1= card(P) =
n+1
X
m=p+1 m−1
p.
k=m−1n+1
p+1=
n
X
k=pk
p
{A1, . . . , An}Ecard(E) =
card(A1) + · · · + card(An)Ai
E A1, . . . , An