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[SPAM] Res: [obm-l] boa de combinatoria
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- Subject: [SPAM] Res: [obm-l] boa de combinatoria
- From: Eduardo Estrada <eestradaitu@xxxxxxxxxxxx>
- Date: Fri, 7 Dec 2007 03:53:48 -0800 (PST)
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--0-1656022518-1197028428=:61224
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Ol=C3=A1, Vit=C3=B3rio,=0A=0AMe parece que a resolu=C3=A7=C3=A3o =C3=A9 a s=
eguinte:=0A=0Aa) Fun=C3=A7=C3=B5es crescentes;=0A=0ABasta que, do contradom=
=C3=ADnio com m elementos, selecionem-se n. A cada sele=C3=A7=C3=A3o, assoc=
ia-se uma =C3=BAnica fun=C3=A7=C3=A3o crescente, e vice-versa. Asim, a resp=
osta =C3=A9 Cm,n. Observe que, quando m<n, o valor obtido =C3=A9 zero, o qu=
e =C3=A9 perfeitamente coerente.=0A=0Ab) Fun=C3=A7=C3=B5es n=C3=A3o decresc=
entes;=0A=0AAnalogamente, o total de fun=C3=A7=C3=B5es decrescentes =C3=A9 =
Cm,n (de fato, observe que, a cada fun=C3=A7=C3=A3o crescente, associa-se u=
ma =C3=BAnica fun=C3=A7=C3=A3o decrescente, e vice-versa). Como o total de =
fun=C3=A7=C3=B5es (de qualquer tipo) =C3=A9 m^n, temos que o valor procurad=
o =C3=A9 m^n - Cm,n.=0A=0AEspero ter ajudado, um abra=C3=A7o!=0AEduardo L. =
Estrada=0A=0A----- Mensagem original ----=0ADe: vitoriogauss <vitoriogauss@=
uol.com.br>=0APara: obm-l <obm-l@xxxxxxxxxxxxxx>=0AEnviadas: Quinta-feira, =
6 de Dezembro de 2007 17:01:58=0AAssunto: [obm-l] boa de combinatoria=0A=0A=
Caros colegas...=0A=0A =0A=0A =0A=0ASeja In =3D {1,2,...,n}, analogamente I=
m, determinar o n=C3=BAmero de fun=C3=A7=C3=B5es f: In --> Im tais que:=0A=
=0A =0A=0A =0A=0Aa) f seja crescente=0A=0A =0A=0Ab) f seja n=C3=A3o-decresc=
ente=0A=0A =0A=0Adesde j=C3=A1 grato....=0A=0A=0A=0A=0A=0A=0A=0A Abra =
sua conta no Yahoo! Mail, o =C3=BAnico sem limite de espa=C3=A7o para armaz=
enamento!=0Ahttp://br.mail.yahoo.com/
--0-1656022518-1197028428=:61224
Content-Type: text/html; charset=utf-8
Content-Transfer-Encoding: quoted-printable
<html><head><style type=3D"text/css"><!-- DIV {margin:0px;} --></style></he=
ad><body><div style=3D"font-family:times new roman,new york,times,serif;fon=
t-size:12pt"><div style=3D"font-family: times new roman,new york,times,seri=
f; font-size: 12pt;">Ol=C3=A1, Vit=C3=B3rio,<br><br>Me parece que a resolu=
=C3=A7=C3=A3o =C3=A9 a seguinte:<br><br>a) Fun=C3=A7=C3=B5es crescentes;<br=
><br>Basta que, do contradom=C3=ADnio com m elementos, selecionem-se n. A c=
ada sele=C3=A7=C3=A3o, associa-se uma =C3=BAnica fun=C3=A7=C3=A3o crescente=
, e vice-versa. Asim, a resposta =C3=A9 Cm,n. Observe que, quando m<n, o=
valor obtido =C3=A9 zero, o que =C3=A9 perfeitamente coerente.<br><br>b) F=
un=C3=A7=C3=B5es n=C3=A3o decrescentes;<br><br>Analogamente, o total de fun=
=C3=A7=C3=B5es <span style=3D"font-style: italic;">decrescentes</span> =C3=
=A9 Cm,n (de fato, observe que, a cada fun=C3=A7=C3=A3o crescente, associa-=
se uma =C3=BAnica fun=C3=A7=C3=A3o decrescente, e vice-versa). Como o total=
de fun=C3=A7=C3=B5es (de qualquer tipo) =C3=A9 m^n, temos que o valor proc=
urado =C3=A9 m^n - Cm,n.<br><br>Espero ter ajudado,
um abra=C3=A7o!<br>Eduardo L. Estrada<br><br><div style=3D"font-family: ti=
mes new roman,new york,times,serif; font-size: 12pt;">----- Mensagem origin=
al ----<br>De: vitoriogauss <vitoriogauss@xxxxxxxxxx><br>Para: obm-l =
<obm-l@xxxxxxxxxxxxxx><br>Enviadas: Quinta-feira, 6 de Dezembro de 20=
07 17:01:58<br>Assunto: [obm-l] boa de combinatoria<br><br><div>Caros coleg=
as...</div>=0A<div> </div>=0A<div> </div>=0A<div>Seja In =3D {1,2=
,...,n}, analogamente Im, determinar o n=C3=BAmero de fun=C3=A7=C3=B5es f: =
In --> Im tais que:</div>=0A<div> </div>=0A<div> </div>=0A<div=
>a) f seja crescente</div>=0A<div> </div>=0A<div>b) f seja n=C3=A3o-de=
crescente</div>=0A<div> </div>=0A<div>desde j=C3=A1 grato....</div>=0A=
</div><br></div></div><br>=0A=0A=0A <hr size=3D1>Abra sua conta no <a =
href=3D"http://br.rd.yahoo.com/mail/taglines/mail/*http://br.mail.yahoo.com=
/">Yahoo! Mail</a>, o =C3=BAnico sem limite de espa=C3=A7o para armazenamen=
to! =0A</body></html>
--0-1656022518-1197028428=:61224--
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