[ ENIBRIS @ 01.09.2005. 20:00 ] @
C(2/3)=3 kombinacije (A,B,C,)= (AB,AC,BC) -> 3 Komb A sad A=(X,Y) ((XY),B,C)=(XB,XC,YB,YC,BC) -> 5 komb Jel postoji neka matematicka formula Hvala unapred |
[ ENIBRIS @ 01.09.2005. 20:00 ] @
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