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Niste module amarate
Posted: Tue Sep 23, 2008 5:42 pm
by Theodor Munteanu
\(
\[
{\rm Aratati ca }\sum\limits_{\sigma \in {\rm S}_{\rm n} ,\varepsilon (\sigma ) = - 1}^{} {{\rm |i - }\sigma {\rm (i)|}} = \sum\limits_{\varepsilon (\sigma ) = 1} {|i - \sigma (i)|} ,\forall n \ge 3,n \in N
\]
\)
Posted: Tue Sep 23, 2008 7:20 pm
by Marius Mainea
|{\( \sigma\in S_n,\epsilon(\sigma)=1,\sigma(i)=k \)}|=|{\( \sigma\in S_n,\epsilon(\sigma)=-1,\sigma(i)=k \)}|=\( \frac{(n-1)!}{2} \)
pentru orice \( k=\overline{1,n} \)
Posted: Tue Sep 23, 2008 7:48 pm
by Theodor Munteanu
Marius Mainea wrote:|{\( \sigma\in S_n,\epsilon(\sigma)=1,\sigma(i)=k \)}|=|{\( \sigma\in S_n,\epsilon(\sigma)=-1,\sigma(i)=k \)}|=\( \frac{(n-1)!}{2} \)
pentru orice \( k=\overline{1,n} \)
Fara suparare da ce sa inteleg eu din asta?
Posted: Wed Sep 24, 2008 9:22 pm
by Marius Mainea
Theodor Munteanu wrote:Marius Mainea wrote:|{\( \sigma\in S_n,\epsilon(\sigma)=1,\sigma(i)=k \)}|=|{\( \sigma\in S_n,\epsilon(\sigma)=-1,\sigma(i)=k \)}|=\( \frac{(n-1)!}{2} \)
pentru orice \( k=\overline{1,n} \)
Fara suparare da ce sa inteleg eu din asta?
\( \sigma(i) \) ia valoarea k de acelasi numar de ori(
\( \frac{(n-1)!}{2} \)) atat pentru termenul din stanga cat si pentru termenul din dreapta.